301. Two identical \(1\,\text{kg}\) blocks are separately released from rest at a common height \(y=5\,\text{m}\) and descend along smooth tracks. Block A reaches an initially undeformed horizontal spring of force constant \(100\,\text{N m}^{-1}\) at height \(y=1\,\text{m}\). Block B reaches an initially undeformed horizontal spring of force constant \(200\,\text{N m}^{-1}\) at height \(y=3\,\text{m}\). Take \(g=10\,\text{m s}^{-2}\). If their speeds immediately before spring contact are \(v_A,v_B\), and their maximum compressions are \(x_A,x_B\), then
ⓐ. \(v_A:v_B=2:1,\;x_A:x_B=\sqrt2:1\)
ⓑ. \(v_A:v_B=\sqrt2:1,\;x_A:x_B=2:1\)
ⓒ. \(v_A:v_B=\sqrt2:1,\;x_A:x_B=\sqrt2:1\)
ⓓ. \(v_A:v_B=1:1,\;x_A:x_B=2:1\)
Correct Answer: \(v_A:v_B=\sqrt2:1,\;x_A:x_B=2:1\)
Explanation: \(\textbf{Speeds before spring contact:}\)
For A, the vertical drop is
\[
5-1=4\,\text{m}
\]
Thus
\[
\frac12v_A^2=(10)(4),
\]
so
\[
v_A^2=80
\]
For B, the vertical drop is
\[
5-3=2\,\text{m}
\]
Hence
\[
v_B^2=40
\]
Therefore
\[
v_A:v_B=\sqrt{80}:\sqrt{40}=\sqrt2:1
\]
\(\textbf{Maximum compression of A:}\)
The available energy is
\[
40\,\text{J}
\]
Thus
\[
40=\frac12(100)x_A^2,
\]
giving
\[
x_A^2=0.8
\]
\(\textbf{Maximum compression of B:}\)
The available energy is
\[
20\,\text{J},
\]
so
\[
20=\frac12(200)x_B^2
\]
Hence
\[
x_B^2=0.2
\]
Therefore
\[
\frac{x_A}{x_B}=\sqrt{\frac{0.8}{0.2}}=2
\]
Both the gravitational drop and the spring stiffness affect the final compression.
\(\textbf{Final answer:}\)
\[
\boxed{v_A:v_B=\sqrt2:1,\qquad x_A:x_B=2:1}
\]
302. A \(1\,\text{kg}\) block moves in a smooth gravity-spring system with total mechanical energy \(20\,\text{J}\). Take \(g=10\,\text{m s}^{-2}\) and spring constant \(k=100\,\text{N m}^{-1}\). Four physically accessible states are specified:
| State | Height \(y\) | Spring deformation \(x\) |
|---|
| P | \(1.0\,\text{m}\) | \(0\) |
| Q | \(0.50\,\text{m}\) | \(0.20\,\text{m}\) |
| R | \(0.20\,\text{m}\) | \(0.40\,\text{m}\) |
| S | \(0\) | \(0.50\,\text{m}\) |
The correct decreasing order of speeds is
ⓐ. \(v_S\gt v_P=v_R\gt v_Q\)
ⓑ. \(v_P\gt v_Q\gt v_R\gt v_S\)
ⓒ. \(v_Q\gt v_R\gt v_P\gt v_S\)
ⓓ. \(v_Q\gt v_P=v_R\gt v_S\)
Correct Answer: \(v_Q\gt v_P=v_R\gt v_S\)
Explanation: At each state,
\[
K=E-(U_g+U_s).
\]
Since the mass is fixed, the speed ranking follows the kinetic-energy ranking.
\(\textbf{State P:}\)
\[
U_g=10\,\text{J},\qquad U_s=0,
\]
so
\[
K_P=10\,\text{J}.
\]
\(\textbf{State Q:}\)
\[
U_g=5\,\text{J},
\]
and
\[
U_s=\frac12(100)(0.20)^2=2\,\text{J}.
\]
Thus
\[
K_Q=20-7=13\,\text{J}.
\]
\(\textbf{State R:}\)
\[
U_g=2\,\text{J},
\]
while
\[
U_s=\frac12(100)(0.40)^2=8\,\text{J}.
\]
Hence
\[
K_R=10\,\text{J}.
\]
\(\textbf{State S:}\)
\[
U_g=0,
\qquad
U_s=\frac12(100)(0.50)^2=12.5\,\text{J}.
\]
Therefore
\[
K_S=7.5\,\text{J}.
\]
Consequently
\[
K_Q\gt K_P=K_R\gt K_S,
\]
and the same ordering holds for the speeds.
\(\textbf{Final answer:}\)
\[
v_Q\gt v_P=v_R\gt v_S.
\]
303. A block of mass \(m\) moves in a smooth system involving gravity and an ideal spring. Between two states, its height decreases from \(y_i\) to \(y_f\), with
\[
h=y_i-y_f\gt 0,
\]
its spring deformation changes from \(x_i\) to \(x_f\), and its speed changes from \(u\) to \(v\).
Statement I:
\[
\frac12mu^2+mgy_i+\frac12kx_i^2
=
\frac12mv^2+mgy_f+\frac12kx_f^2.
\]
Statement II:
\[
\frac12m(v^2-u^2)
=
mgh+\frac12k(x_i^2-x_f^2).
\]
Statement III: If the height decreases and the magnitude of spring deformation also decreases, both potential-energy forms decrease, so the kinetic energy must increase.
Statement IV: One may retain both \(mgy_i\) and \(mgy_f\) in the conservation equation and additionally add \(+mgh\) as a separate gravitational-work term without changing the energy accounting.
Which of the following is correct?
ⓐ. Statements I and II only
ⓑ. Statements I, II and III only
ⓒ. Statements II, III and IV only
ⓓ. Statements I, III and IV only
Correct Answer: Statements I, II and III only
Explanation: Statement I is the direct mechanical-energy conservation equation for the stated smooth system.
It consistently includes kinetic energy, gravitational potential energy and spring potential energy at each endpoint.
Rearranging Statement I gives
\[
\frac12m(v^2-u^2)
=
mg(y_i-y_f)
+
\frac12k(x_i^2-x_f^2).
\]
Since
\[
h=y_i-y_f,
\]
this is exactly Statement II.
Statement III is also correct.
A decrease in height lowers \(U_g\), while a decrease in \(|x|\) lowers
\[
U_s=\frac12kx^2.
\]
If no other work removes energy, the combined decrease in potential energy appears as increased kinetic energy.
Statement IV is incorrect because it counts the gravitational energy transfer twice.
One may describe gravity either through the potential-energy difference or through gravitational work in an appropriate alternative formulation, but not add both representations of the same transfer in one energy balance.
Hence Statements I, II and III are correct.
304.
A \(1\,\text{kg}\) block slides down a smooth incline with \(\sin\theta=0.6\) and reaches the natural-length position of an ideal spring with speed \(4\,\text{m s}^{-1}\). The spring lies along the incline and has force constant \(40\,\text{N m}^{-1}\). The block compresses the spring, reverses, and is pushed back upward. On returning to natural length it loses contact with the spring and continues up the smooth incline. Take \(g=10\,\text{m s}^{-2}\).
What are the maximum spring compression, the speed when the block returns to natural length, and the maximum vertical height reached above the natural-length position?
ⓐ. \(0.50\,\text{m},\;4\,\text{m s}^{-1},\;0.80\,\text{m}\)
ⓑ. \(0.80\,\text{m},\;2\sqrt2\,\text{m s}^{-1},\;0.40\,\text{m}\)
ⓒ. \(0.80\,\text{m},\;4\,\text{m s}^{-1},\;0.80\,\text{m}\)
ⓓ. \(1.0\,\text{m},\;4\,\text{m s}^{-1},\;1.20\,\text{m}\)
Correct Answer: \(0.80\,\text{m},\;4\,\text{m s}^{-1},\;0.80\,\text{m}\)
Explanation: \(\textbf{Compression stage:}\)
At natural length the initial kinetic energy is
\[
K_i=\frac12(1)(4^2)=8\,\text{J}
\]
If the maximum compression is \(X\), the block descends vertically by
\[
X\sin\theta=0.6X
\]
Gravity therefore releases
\[
mg(0.6X)=6X
\]
joules during the compression.
At maximum compression,
\[
K=0,
\]
so
\[
8+6X=\frac12(40)X^2
\]
Thus
\[
8+6X=20X^2
\]
Rearranging,
\[
20X^2-6X-8=0
\]
The physical positive root is
\[
X=0.80\,\text{m}
\]
\(\textbf{Return to natural length:}\)
The system returns to the same spring deformation and the same height at which the compression stage began.
With only conservative gravity and spring interactions,
\[
K_{\text{return}}=8\,\text{J}
\]
Therefore the returning speed has the same magnitude:
\[
v=4\,\text{m s}^{-1}
\]
\(\textbf{Climb after leaving the spring:}\)
The remaining kinetic energy converts completely into gravitational potential energy:
\[
mgh=8
\]
Hence
\[
(1)(10)h=8,
\]
giving
\[
h=0.80\,\text{m}
\]
\(\textbf{Final answer:}\)
\[
\boxed{0.80\,\text{m},\;4\,\text{m s}^{-1},\;0.80\,\text{m}}
\]
305. A \(2\,\text{kg}\) block moves in a smooth system involving gravity and an ideal spring of unknown force constant \(k\). Take \(g=10\,\text{m s}^{-2}\). At State A,
\[
y_A=1.0\,\text{m},\qquad x_A=0.10\,\text{m},\qquad v_A=3\,\text{m s}^{-1}.
\]
At State B,
\[
y_B=0.40\,\text{m},\qquad x_B=0.30\,\text{m},\qquad v_B=\sqrt5\,\text{m s}^{-1}.
\]
In a second trial, the spring and the two configurations A and B are unchanged, but the block mass is \(4\,\text{kg}\). The \(4\,\text{kg}\) block starts at A with the same speed \(3\,\text{m s}^{-1}\). What are \(k\) and its speed on reaching B?
ⓐ. \(200\,\text{N m}^{-1},\;\sqrt{13}\,\text{m s}^{-1}\)
ⓑ. \(400\,\text{N m}^{-1},\;\sqrt{13}\,\text{m s}^{-1}\)
ⓒ. \(400\,\text{N m}^{-1},\;\sqrt{26}\,\text{m s}^{-1}\)
ⓓ. \(800\,\text{N m}^{-1},\;\sqrt5\,\text{m s}^{-1}\)
Correct Answer: \(400\,\text{N m}^{-1},\;\sqrt{13}\,\text{m s}^{-1}\)
Explanation: Mechanical energy is conserved in each smooth trial.
\(\textbf{Use the first trial to determine }k\textbf{:}\)
At A,
\[
K_A=\frac12(2)(3^2)=9\,\text{J},
\]
\[
U_{g,A}=(2)(10)(1)=20\,\text{J},
\]
and
\[
U_{s,A}=\frac12k(0.10)^2=0.005k
\]
At B,
\[
K_B=\frac12(2)(\sqrt5)^2=5\,\text{J},
\]
\[
U_{g,B}=(2)(10)(0.40)=8\,\text{J},
\]
and
\[
U_{s,B}=\frac12k(0.30)^2=0.045k
\]
Hence
\[
9+20+0.005k=5+8+0.045k
\]
Therefore
\[
16=0.04k,
\]
so
\[
k=400\,\text{N m}^{-1}
\]
\(\textbf{Second trial with mass }4\,\text{kg}\textbf{:}\)
At A,
\[
E= \frac12(4)(3^2)+(4)(10)(1)+\frac12(400)(0.10)^2
\]
\[
E=18+40+2=60\,\text{J}
\]
At B,
\[
U_g=(4)(10)(0.40)=16\,\text{J},
\]
\[
U_s=\frac12(400)(0.30)^2=18\,\text{J}
\]
Thus
\[
K_B=60-16-18=26\,\text{J}
\]
Therefore
\[
26=\frac12(4)v^2=2v^2,
\]
giving
\[
v=\sqrt{13}\,\text{m s}^{-1}
\]
\(\textbf{Final answer:}\)
\[
\boxed{k=400\,\text{N m}^{-1},\qquad v_B=\sqrt{13}\,\text{m s}^{-1}}
\]
306. A \(1\,\text{kg}\) block moves in a smooth gravity-spring system with \(g=10\,\text{m s}^{-2}\) and spring constant \(200\,\text{N m}^{-1}\). Four states are recorded using the same gravitational and spring potential-energy references.
| State | Height \(y\) | Spring deformation \(x\) | Kinetic energy \(K\) |
|---|
| P | \(1.0\,\text{m}\) | \(0\) | \(10\,\text{J}\) |
| Q | \(0.50\,\text{m}\) | \(0.20\,\text{m}\) | \(11\,\text{J}\) |
| R | \(0.20\,\text{m}\) | \(0.40\,\text{m}\) | \(2\,\text{J}\) |
| S | \(0\) | \(0.30\,\text{m}\) | \(12\,\text{J}\) |
If P, Q and R belong to one mechanically isolated trajectory, which state is inconsistent with that same conserved mechanical energy?
ⓐ. P
ⓑ. Q
ⓒ. R
ⓓ. S
Correct Answer: S
Explanation: For each state,
\[
E=K+mgy+\frac12kx^2.
\]
At P,
\[
E_P=10+10+0=20\,\text{J}.
\]
At Q,
\[
E_Q=11+5+\frac12(200)(0.20)^2.
\]
\[
E_Q=11+5+4=20\,\text{J}.
\]
At R,
\[
E_R=2+2+\frac12(200)(0.40)^2.
\]
\[
E_R=2+2+16=20\,\text{J}.
\]
Thus P, Q and R are mutually consistent with
\[
E=20\,\text{J}.
\]
At S,
\[
E_S=12+0+\frac12(200)(0.30)^2.
\]
\[
E_S=12+9=21\,\text{J}.
\]
Therefore S cannot belong to the same conservative trajectory without an additional energy transfer.
307.
A \(1\,\text{kg}\) block is initially held against Spring 1, of force constant \(200\,\text{N m}^{-1}\), compressed by \(0.20\,\text{m}\) on a smooth horizontal surface. It is released and loses contact with Spring 1 at its natural length. The block then follows a smooth track upward and reaches the natural-length position of Spring 2 at a vertical height \(0.10\,\text{m}\) above its starting level. Spring 2 has force constant \(100\,\text{N m}^{-1}\) and lies along a smooth incline with \(\sin\theta=1/2\). The block compresses Spring 2 while moving farther upward. Take \(g=10\,\text{m s}^{-2}\).
What are the maximum compression of Spring 2 and the block's speed when Spring 2 is compressed by \(0.10\,\text{m}\)?
ⓐ. \(0.20\,\text{m},\;\sqrt6\,\text{m s}^{-1}\)
ⓑ. \(0.10\,\text{m},\;2\sqrt2\,\text{m s}^{-1}\)
ⓒ. \(0.30\,\text{m},\;\sqrt2\,\text{m s}^{-1}\)
ⓓ. \(0.20\,\text{m},\;2\,\text{m s}^{-1}\)
Correct Answer: \(0.20\,\text{m},\;2\,\text{m s}^{-1}\)
Explanation: The initial energy is entirely stored in Spring 1:
\[
E=\frac12(200)(0.20)^2=4\,\text{J}
\]
No dissipative force acts, so this total energy remains available throughout the later motion.
\(\textbf{Maximum compression of Spring 2:}\)
If its compression is \(X\), the block has risen to height
\[
0.10+\frac X2
\]
At maximum compression,
\[
K=0
\]
Hence
\[
4=(1)(10)\left(0.10+\frac X2\right)+\frac12(100)X^2
\]
Thus
\[
4=1+5X+50X^2
\]
Therefore
\[
50X^2+5X-3=0
\]
The positive root is
\[
X=0.20\,\text{m}
\]
\(\textbf{At compression }0.10\,\text{m}\textbf{:}\)
The height is
\[
0.10+\frac{0.10}{2}=0.15\,\text{m}
\]
Thus
\[
U_g=(1)(10)(0.15)=1.5\,\text{J},
\]
and
\[
U_s=\frac12(100)(0.10)^2=0.5\,\text{J}
\]
Hence
\[
K=4-1.5-0.5=2\,\text{J}
\]
Therefore
\[
\frac12v^2=2,
\]
so
\[
v=2\,\text{m s}^{-1}
\]
\(\textbf{Final answer:}\)
\[
\boxed{X_{\max}=0.20\,\text{m},\qquad v=2\,\text{m s}^{-1}}
\]
308. A \(1\,\text{kg}\) block is attached to a vertical ideal spring of force constant \(50\,\text{N m}^{-1}\). Let \(x\) denote downward displacement from the spring's natural length, and take gravitational potential energy to be zero at \(x=0\). At \(x=+0.20\,\text{m}\), the block is moving downward with speed \(2\,\text{m s}^{-1}\). Which set gives its first lower turning position, its later upper turning position, and its speed when it passes through \(x=0\)?
ⓐ. \(\dfrac{1+\sqrt2}{5}\,\text{m},\;\dfrac{1-\sqrt2}{5}\,\text{m},\;\sqrt2\,\text{m s}^{-1}\)
ⓑ. \(\dfrac{1-\sqrt2}{5}\,\text{m},\;\dfrac{1+\sqrt2}{5}\,\text{m},\;\sqrt2\,\text{m s}^{-1}\)
ⓒ. \(\dfrac{2+\sqrt2}{5}\,\text{m},\;-\dfrac{\sqrt2}{5}\,\text{m},\;2\,\text{m s}^{-1}\)
ⓓ. \(\dfrac{1+\sqrt2}{5}\,\text{m},\;0,\;1\,\text{m s}^{-1}\)
Correct Answer: \(\dfrac{1+\sqrt2}{5}\,\text{m},\;\dfrac{1-\sqrt2}{5}\,\text{m},\;\sqrt2\,\text{m s}^{-1}\)
Explanation: With downward \(x\) positive,
\[
U_g=-mgx,
\]
while
\[
U_s=\frac12kx^2
\]
\(\textbf{Mechanical energy at the stated point:}\)
\[
K=\frac12(1)(2^2)=2\,\text{J},
\]
\[
U_s=\frac12(50)(0.20)^2=1\,\text{J},
\]
and
\[
U_g=-(1)(10)(0.20)=-2\,\text{J}
\]
Thus
\[
E=2+1-2=1\,\text{J}
\]
At either turning point,
\[
K=0
\]
Therefore
\[
25x^2-10x=1
\]
Hence
\[
25x^2-10x-1=0
\]
The two roots are
\[
x=\frac{1\pm\sqrt2}{5}
\]
Because the block is initially moving downward, the first turning point is the positive root
\[
x_{\text{lower}}=\frac{1+\sqrt2}{5}\,\text{m}
\]
The later upper turning point is
\[
x_{\text{upper}}=\frac{1-\sqrt2}{5}\,\text{m},
\]
which is negative.
At the natural-length position,
\[
x=0,
\]
both potential-energy terms are zero.
Thus
\[
K=E=1\,\text{J}
\]
Therefore
\[
\frac12v^2=1,
\]
giving
\[
v=\sqrt2\,\text{m s}^{-1}
\]
\(\textbf{Final answer:}\)
\[
\boxed{\frac{1+\sqrt2}{5}\,\text{m},\;\frac{1-\sqrt2}{5}\,\text{m},\;\sqrt2\,\text{m s}^{-1}}
\]
309. Consider the following statements about mechanical-energy conservation.
Statement I: Shifting the zero of gravitational or spring potential energy by an additive constant can change the numerical value assigned to total mechanical energy without changing any predicted motion.
Statement II: A turning point is identified by zero instantaneous kinetic energy, but the net force there need not be zero.
Statement III: The mere presence of a normal reaction necessarily destroys mechanical-energy conservation.
Statement IV: If gravity is already represented through \(mgy\) in a mechanical-energy equation, adding gravitational work separately for the same displacement double-counts the same energy transfer.
Which of the following is correct?
ⓐ. Statements I and II only
ⓑ. Statements I, II and IV only
ⓒ. Statements II, III and IV only
ⓓ. Statements I, III and IV only
Correct Answer: Statements I, II and IV only
Explanation: Statement I is correct because the zero of potential energy is conventional.
Adding a constant to all values of a potential energy changes the numerical value of \(K+U\) but leaves every potential-energy difference unchanged.
Consequently speeds, turning points and forces are unaffected.
Statement II is also correct.
A turning point requires
\[
K=0,
\]
but the acceleration may be non-zero there, allowing the body to reverse direction.
Statement III is incorrect because a normal reaction can do zero work, as on many smooth fixed tracks.
Its presence alone does not determine whether mechanical energy is conserved.
Statement IV is correct because gravitational potential-energy change and gravitational work are alternative accounting descriptions of the same conservative interaction.
Using both in the same balance for the same transfer counts gravity twice.
310.
A \(1\,\text{kg}\) block is attached to an ideal spring of force constant \(100\,\text{N m}^{-1}\) lying along a smooth incline with \(\sin\theta=1/2\). The spring is initially undeformed. The block starts at the natural-length position moving down the incline at \(2\,\text{m s}^{-1}\). A constant external force of \(5\,\text{N}\) acts down the incline only until the spring is stretched by \(0.20\,\text{m}\), where the external force is removed. Take \(g=10\,\text{m s}^{-2}\).
What are the speed at the instant the external force is removed, the subsequent maximum spring deformation, and the speed when the block later returns to the natural-length position?
ⓐ. \(\sqrt2\,\text{m s}^{-1},\;0.20\,\text{m},\;2\,\text{m s}^{-1}\)
ⓑ. \(2\,\text{m s}^{-1},\;0.30\,\text{m},\;\sqrt6\,\text{m s}^{-1}\)
ⓒ. \(2\sqrt2\,\text{m s}^{-1},\;0.30\,\text{m},\;2\,\text{m s}^{-1}\)
ⓓ. \(2\,\text{m s}^{-1},\;0.40\,\text{m},\;\sqrt3\,\text{m s}^{-1}\)
Correct Answer: \(2\,\text{m s}^{-1},\;0.30\,\text{m},\;\sqrt6\,\text{m s}^{-1}\)
Explanation: Take \(x\) positive down the incline from the spring's natural length.
\(\textbf{Motion while the external force acts:}\)
The initial kinetic energy is
\[
K_i=\frac12(1)(2^2)=2\,\text{J}
\]
Over \(0.20\,\text{m}\), gravity does
\[
W_g=mg\sin\theta\,x=(1)(10)\left(\frac12\right)(0.20)=1\,\text{J}
\]
The external force does
\[
W_{\text{ext}}=(5)(0.20)=1\,\text{J}
\]
The spring does
\[
W_s=-\frac12(100)(0.20)^2=-2\,\text{J}
\]
Thus
\[
\Delta K=1+1-2=0
\]
Therefore the speed at removal remains
\[
2\,\text{m s}^{-1}
\]
\(\textbf{Mechanical energy after removal:}\)
Using
\[
U_g=-mgx\sin\theta=-5x,
\]
the energy at \(x=0.20\,\text{m}\) is
\[
E=2+\frac12(100)(0.20)^2-5(0.20)
\]
\[
E=2+2-1=3\,\text{J}
\]
At the next turning point \(X\),
\[
50X^2-5X=3
\]
The positive solution is
\[
X=0.30\,\text{m}
\]
When the block later returns to \(x=0\),
\[
U_g=U_s=0,
\]
so
\[
K=3\,\text{J}
\]
Therefore
\[
\frac12v^2=3,
\]
giving
\[
v=\sqrt6\,\text{m s}^{-1}
\]
\(\textbf{Final answer:}\)
\[
\boxed{2\,\text{m s}^{-1},\;0.30\,\text{m},\;\sqrt6\,\text{m s}^{-1}}
\]
311. A \(1\,\text{kg}\) block reaches an initially undeformed spring of force constant \(36\,\text{N m}^{-1}\) with speed \(2\,\text{m s}^{-1}\). Three separate smooth arrangements are considered. In H, the spring is horizontal. In D, the spring lies along an incline with \(\sin\theta=1/2\) and the block moves downhill while compressing it. In U, the same incline is used but the block moves uphill while compressing the spring. Take \(g=10\,\text{m s}^{-2}\). The maximum compressions \(x_H,x_D,x_U\) are
ⓐ. \(\dfrac12\,\text{m},\;\dfrac13\,\text{m},\;\dfrac29\,\text{m}\)
ⓑ. \(\dfrac13\,\text{m},\;\dfrac29\,\text{m},\;\dfrac12\,\text{m}\)
ⓒ. \(\dfrac13\,\text{m},\;\dfrac12\,\text{m},\;\dfrac29\,\text{m}\)
ⓓ. \(\dfrac29\,\text{m},\;\dfrac12\,\text{m},\;\dfrac13\,\text{m}\)
Correct Answer: \(\dfrac13\,\text{m},\;\dfrac12\,\text{m},\;\dfrac29\,\text{m}\)
Explanation: The kinetic energy at first spring contact is
\[
K_i=\frac12(1)(2^2)=2\,\text{J}
\]
\(\textbf{Horizontal spring:}\)
At maximum compression,
\[
2=\frac12(36)x_H^2
\]
Thus
\[
18x_H^2=2,
\]
giving
\[
x_H=\frac13\,\text{m}
\]
\(\textbf{Compression while moving downhill:}\)
Gravity supplies additional energy:
\[
2+mg\sin\theta\,x_D=\frac12kx_D^2
\]
Therefore
\[
2+5x_D=18x_D^2
\]
The positive solution is
\[
x_D=\frac12\,\text{m}
\]
\(\textbf{Compression while moving uphill:}\)
Now gravitational potential energy increases:
\[
2=18x_U^2+5x_U
\]
The positive solution is
\[
x_U=\frac29\,\text{m}
\]
Thus the downhill spring compresses most because gravity assists the motion, whereas the uphill spring compresses least because gravity competes with the initial kinetic energy.
\(\textbf{Final answer:}\)
\[
\boxed{x_H=\frac13\,\text{m},\quad x_D=\frac12\,\text{m},\quad x_U=\frac29\,\text{m}}
\]
312.
A \(1\,\text{kg}\) block on a smooth horizontal table is connected by a light inextensible string over a frictionless pulley to a hanging \(1\,\text{kg}\) block. The table block is also attached to a horizontal ideal spring of force constant \(50\,\text{N m}^{-1}\), initially undeformed. The system is released from rest. As the hanging block descends by distance \(x\), the spring stretches by the same distance. Take \(g=10\,\text{m s}^{-2}\).
What are the maximum spring extension, the common speed when \(x=0.20\,\text{m}\), and the string tension at that instant?
ⓐ. \(0.20\,\text{m},\;\sqrt2\,\text{m s}^{-1},\;5\,\text{N}\)
ⓑ. \(0.40\,\text{m},\;1\,\text{m s}^{-1},\;10\,\text{N}\)
ⓒ. \(0.40\,\text{m},\;2\,\text{m s}^{-1},\;5\,\text{N}\)
ⓓ. \(0.50\,\text{m},\;1\,\text{m s}^{-1},\;15\,\text{N}\)
Correct Answer: \(0.40\,\text{m},\;1\,\text{m s}^{-1},\;10\,\text{N}\)
Explanation: Use the two blocks plus spring plus Earth as the mechanical system.
\(\textbf{Maximum extension:}\)
When the spring extension is \(x\), the hanging block has lost gravitational potential energy
\[
mgx=10x
\]
At the maximum extension \(X\), both blocks are instantaneously at rest:
\[
10X=\frac12(50)X^2
\]
Hence
\[
10X=25X^2,
\]
so
\[
X=0.40\,\text{m}
\]
\(\textbf{Speed at }x=0.20\,\text{m}\textbf{:}\)
The gravitational energy released is
\[
(10)(0.20)=2\,\text{J}
\]
The spring stores
\[
\frac12(50)(0.20)^2=1\,\text{J}
\]
The remaining \(1\,\text{J}\) is the total kinetic energy of the two \(1\,\text{kg}\) blocks:
\[
K_{\text{tot}}=\frac12v^2+\frac12v^2=v^2
\]
Therefore
\[
v=1\,\text{m s}^{-1}
\]
\(\textbf{Tension at this extension:}\)
At \(x=0.20\,\text{m}\), the spring force is
\[
kx=(50)(0.20)=10\,\text{N}
\]
The hanging weight is also \(10\,\text{N}\), so the net driving force on the complete moving system is zero at that instant.
Thus
\[
a=0
\]
For the hanging block,
\[
mg-T=0,
\]
giving
\[
T=10\,\text{N}
\]
\(\textbf{Final answer:}\)
\[
\boxed{0.40\,\text{m},\;1\,\text{m s}^{-1},\;10\,\text{N}}
\]
313. An external agent very slowly moves a \(1\,\text{kg}\) block in a conservative gravity-spring arrangement from a state with \(y=0,\;x=0\) to a state with
\[
y=0.50\,\text{m},\qquad x=0.20\,\text{m}.
\]
The spring constant is \(100\,\text{N m}^{-1}\), and \(g=10\,\text{m s}^{-2}\). The block is then released from rest. Later it reaches
\[
y=0.20\,\text{m},\qquad x=0.10\,\text{m}.
\]
What are the work done by the external agent during the slow stage and the block's speed at the later state?
ⓐ. \(5\,\text{J},\;2\,\text{m s}^{-1}\)
ⓑ. \(7\,\text{J},\;3\,\text{m s}^{-1}\)
ⓒ. \(7\,\text{J},\;\sqrt7\,\text{m s}^{-1}\)
ⓓ. \(9\,\text{J},\;3\,\text{m s}^{-1}\)
Correct Answer: \(7\,\text{J},\;3\,\text{m s}^{-1}\)
Explanation: Because the first stage is very slow, the change in kinetic energy during that stage is negligible.
\(\textbf{Work supplied by the external agent:}\)
The gravitational potential-energy increase is
\[
\Delta U_g=(1)(10)(0.50)=5\,\text{J}
\]
The spring potential-energy increase is
\[
\Delta U_s=\frac12(100)(0.20)^2=2\,\text{J}
\]
Thus
\[
W_{\text{ext}}=5+2=7\,\text{J}
\]
The released system therefore begins its free motion with mechanical energy
\[
E=7\,\text{J}
\]
\(\textbf{Later state:}\)
The gravitational potential energy is
\[
U_g=(1)(10)(0.20)=2\,\text{J}
\]
The spring potential energy is
\[
U_s=\frac12(100)(0.10)^2=0.5\,\text{J}
\]
Hence
\[
K=7-2-0.5=4.5\,\text{J}
\]
Therefore
\[
\frac12v^2=4.5,
\]
so
\[
v=3\,\text{m s}^{-1}
\]
\(\textbf{Final answer:}\)
\[
\boxed{W_{\text{ext}}=7\,\text{J},\qquad v=3\,\text{m s}^{-1}}
\]
314. Match List I with List II for the indicated choice of system in each case. All surfaces mentioned are smooth and fixed.
| List I | Chosen system and situation | List II | Correct energy statement |
|---|
| P | Block + Earth; motion under gravity | 1 | \(K+U_g=\text{constant}\) |
| Q | Block + spring; horizontal spring motion | 2 | \(K+U_s=\text{constant}\) |
| R | Block + Earth + spring; gravity-spring motion | 3 | \(K+U_g+U_s=\text{constant}\) |
| S | Block alone; gravity acts externally | 4 | \(\Delta K=W_g\) |
Choose the correct matching.
ⓐ. P-3, Q-1, R-4, S-2
ⓑ. P-4, Q-2, R-1, S-3
ⓒ. P-1, Q-3, R-2, S-4
ⓓ. P-1, Q-2, R-3, S-4
Correct Answer: P-1, Q-2, R-3, S-4
Explanation: For P, Earth is included with the block, so the gravitational interaction is represented by \(U_g\).
Therefore
\[
K+U_g=\text{constant}.
\]
For Q, the spring interaction is internal to the selected block-spring system.
Hence
\[
K+U_s=\text{constant}.
\]
For R, both conservative interactions are included within the selected system.
Thus
\[
K+U_g+U_s=\text{constant}.
\]
For S, Earth is outside the selected block-only system.
Gravity is then treated as an external force whose work changes the block's kinetic energy:
\[
\Delta K=W_g.
\]
The different equations do not contradict one another; they reflect different but consistent system boundaries.
Therefore the correct matching is
\[
\text{P-1,\ Q-2,\ R-3,\ S-4}.
\]
315. A \(1\,\text{kg}\) block reaches an initially undeformed spring lying along a smooth incline. Both the spring constant \(k\) and the incline angle \(\theta\) are unknown. Take \(g=10\,\text{m s}^{-2}\). In Trial I, the block reaches the spring moving downhill with speed \(\sqrt6\,\text{m s}^{-1}\) and compresses it by a maximum of \(0.30\,\text{m}\). In Trial II, it reaches the same spring downhill with speed \(2\sqrt3\,\text{m s}^{-1}\) and the maximum compression is \(0.40\,\text{m}\). If in Trial III it reaches the uncompressed spring from rest, what are \(k\), \(\sin\theta\), and the non-zero maximum compression in Trial III?
ⓐ. \(50\,\text{N m}^{-1},\;1/2,\;0.20\,\text{m}\)
ⓑ. \(100\,\text{N m}^{-1},\;3/5,\;0.12\,\text{m}\)
ⓒ. \(100\,\text{N m}^{-1},\;1/2,\;0.10\,\text{m}\)
ⓓ. \(200\,\text{N m}^{-1},\;1/2,\;0.05\,\text{m}\)
Correct Answer: \(100\,\text{N m}^{-1},\;1/2,\;0.10\,\text{m}\)
Explanation: Let
\[
s=\sin\theta
\]
During downhill compression, gravity supplies additional energy.
\(\textbf{Trial I:}\)
The initial kinetic energy is
\[
K_1=\frac12(1)(\sqrt6)^2=3\,\text{J}
\]
At maximum compression,
\[
3+(10s)(0.30)=\frac12k(0.30)^2
\]
Thus
\[
3+3s=0.045k
\]
\(\textbf{Trial II:}\)
The contact kinetic energy is
\[
K_2=\frac12(1)(2\sqrt3)^2=6\,\text{J}
\]
Therefore
\[
6+(10s)(0.40)=\frac12k(0.40)^2
\]
Hence
\[
6+4s=0.08k
\]
Solving the two simultaneous relations gives
\[
s=\frac12,
\qquad
k=100\,\text{N m}^{-1}
\]
\(\textbf{Trial III from rest:}\)
Let the non-zero maximum compression be \(X\).
Then
\[
mg\sin\theta\,X=\frac12kX^2
\]
Thus
\[
5X=50X^2
\]
For the non-zero solution,
\[
X=0.10\,\text{m}
\]
\(\textbf{Final answer:}\)
\[
\boxed{k=100\,\text{N m}^{-1},\qquad \sin\theta=\frac12,\qquad X=0.10\,\text{m}}
\]
316. A \(1\,\text{kg}\) block initially rests at \(y=0\) with an ideal spring undeformed. The spring constant is \(100\,\text{N m}^{-1}\). A motor transfers mechanical energy to the block-Earth-spring system at constant power \(60\,\text{W}\) for \(0.20\,\text{s}\), after which the motor is disconnected. At disconnection,
\[
y=0.50\,\text{m},\qquad x=0.20\,\text{m}.
\]
The subsequent motion is smooth and conservative. Later,
\[
y=0.20\,\text{m},\qquad x=0.30\,\text{m}.
\]
Take \(g=10\,\text{m s}^{-2}\). Which set gives the speed at disconnection, the mechanical energy available after disconnection, and the later speed?
ⓐ. \(\sqrt{10}\,\text{m s}^{-1},\;12\,\text{J},\;\sqrt{11}\,\text{m s}^{-1}\)
ⓑ. \(\sqrt6\,\text{m s}^{-1},\;12\,\text{J},\;\sqrt{10}\,\text{m s}^{-1}\)
ⓒ. \(\sqrt{10}\,\text{m s}^{-1},\;7\,\text{J},\;3\,\text{m s}^{-1}\)
ⓓ. \(2\sqrt3\,\text{m s}^{-1},\;12\,\text{J},\;\sqrt7\,\text{m s}^{-1}\)
Correct Answer: \(\sqrt{10}\,\text{m s}^{-1},\;12\,\text{J},\;\sqrt{11}\,\text{m s}^{-1}\)
Explanation: The motor supplies mechanical work
\[
W_{\text{motor}}=P\Delta t
\]
Therefore
\[
W_{\text{motor}}=(60)(0.20)=12\,\text{J}
\]
The system began with zero kinetic and zero chosen potential energies, so its mechanical energy at disconnection is
\[
E=12\,\text{J}
\]
\(\textbf{State at disconnection:}\)
The gravitational potential energy is
\[
U_g=(1)(10)(0.50)=5\,\text{J}
\]
The spring potential energy is
\[
U_s=\frac12(100)(0.20)^2=2\,\text{J}
\]
Thus
\[
K=12-5-2=5\,\text{J}
\]
Hence
\[
\frac12v^2=5,
\]
giving
\[
v=\sqrt{10}\,\text{m s}^{-1}
\]
\(\textbf{Later conservative state:}\)
\[
U_g=(1)(10)(0.20)=2\,\text{J},
\]
and
\[
U_s=\frac12(100)(0.30)^2=4.5\,\text{J}
\]
Therefore
\[
K=12-2-4.5=5.5\,\text{J}
\]
Thus
\[
\frac12v^2=5.5,
\]
so
\[
v=\sqrt{11}\,\text{m s}^{-1}
\]
\(\textbf{Final answer:}\)
\[
\boxed{\sqrt{10}\,\text{m s}^{-1},\;12\,\text{J},\;\sqrt{11}\,\text{m s}^{-1}}
\]
317.
A \(1\,\text{kg}\) block is released from rest \(0.15\,\text{m}\) vertically above the natural-length position of a vertical spring of force constant \(100\,\text{N m}^{-1}\). It falls freely, contacts the spring, and continues downward while compressing it. Take \(g=10\,\text{m s}^{-2}\).
What are the maximum compression, the magnitude and direction of the net force at maximum compression, and the speed when the compression equals the static force-balance value \(mg/k\)?
ⓐ. \(0.20\,\text{m},\;10\,\text{N}\) upward, \(\sqrt2\,\text{m s}^{-1}\)
ⓑ. \(0.30\,\text{m},\;20\,\text{N}\) upward, \(2\,\text{m s}^{-1}\)
ⓒ. \(0.30\,\text{m},\;30\,\text{N}\) upward, \(\sqrt3\,\text{m s}^{-1}\)
ⓓ. \(0.40\,\text{m},\;20\,\text{N}\) downward, \(2\,\text{m s}^{-1}\)
Correct Answer: \(0.30\,\text{m},\;20\,\text{N}\) upward, \(2\,\text{m s}^{-1}\)
Explanation: The block has gained kinetic energy
\[
mgh=(1)(10)(0.15)=1.5\,\text{J}
\]
by the instant it first contacts the spring.
Let \(x\) be downward compression from the natural-length position.
\(\textbf{Maximum compression:}\)
During compression, gravity continues doing positive work.
At the turning point,
\[
1.5+10X=\frac12(100)X^2
\]
Thus
\[
1.5+10X=50X^2
\]
The positive solution is
\[
X=0.30\,\text{m}
\]
\(\textbf{Net force at the turning point:}\)
The spring force is
\[
F_s=kX=(100)(0.30)=30\,\text{N}
\]
upward.
The weight is \(10\,\text{N}\) downward.
Hence the net force is
\[
30-10=20\,\text{N}
\]
upward.
Thus zero speed at the turning point does not mean zero force.
\(\textbf{At the force-balance compression:}\)
\[
x_{\text{eq}}=\frac{mg}{k}=\frac{10}{100}=0.10\,\text{m}
\]
Using energy from first contact,
\[
K+50(0.10)^2-10(0.10)=1.5
\]
Therefore
\[
K+0.5-1=1.5,
\]
so
\[
K=2\,\text{J}
\]
Hence
\[
\frac12v^2=2,
\]
giving
\[
v=2\,\text{m s}^{-1}
\]
\(\textbf{Final answer:}\)
\[
\boxed{0.30\,\text{m},\;20\,\text{N upward},\;2\,\text{m s}^{-1}}
\]
318. A \(1\,\text{kg}\) block moves in a smooth gravity-spring system with \(g=10\,\text{m s}^{-2}\) and \(k=100\,\text{N m}^{-1}\). At State A,
\[
y_A=0.60\,\text{m},\qquad x_A=0.30\,\text{m},\qquad v_A=2\,\text{m s}^{-1}.
\]
At State B,
\[
y_B=0.20\,\text{m},\qquad x_B=0.10\,\text{m}.
\]
Four students propose methods for finding \(v_B\).
Student P:
\[
K_A+U_{g,A}+U_{s,A}=K_B+U_{g,B}+U_{s,B}.
\]
Student Q:
\[
K_B-K_A=W_g+W_s.
\]
Student R:
\[
K_A+U_{g,A}+U_{s,A}+W_g+W_s=K_B+U_{g,B}+U_{s,B}.
\]
Student S:
\[
K_B-K_A=-(\Delta U_g+\Delta U_s).
\]
Which students use consistent energy accounting, and what is \(v_B\)?
ⓐ. P and Q only; \(v_B=2\sqrt5\,\text{m s}^{-1}\)
ⓑ. Q, R and S; \(v_B=4\,\text{m s}^{-1}\)
ⓒ. P, Q and S; \(v_B=2\sqrt5\,\text{m s}^{-1}\)
ⓓ. P and R only; \(v_B=2\sqrt3\,\text{m s}^{-1}\)
Correct Answer: P, Q and S; \(v_B=2\sqrt5\,\text{m s}^{-1}\)
Explanation: Student P correctly applies mechanical-energy conservation.
Student Q correctly uses the work-energy theorem with the works of gravity and the spring.
Student S is algebraically equivalent because conservative-force work equals minus the corresponding potential-energy change.
Student R is incorrect because it includes \(U_g\) and \(U_s\) and then adds the same conservative-force works again, thereby double-counting those energy transfers.
\(\textbf{Calculate the endpoint energies:}\)
At A,
\[
K_A=\frac12(1)(2^2)=2\,\text{J}
\]
\[
U_{g,A}=(1)(10)(0.60)=6\,\text{J},
\]
and
\[
U_{s,A}=\frac12(100)(0.30)^2=4.5\,\text{J}
\]
Thus
\[
E=12.5\,\text{J}
\]
At B,
\[
U_{g,B}=2\,\text{J},
\]
and
\[
U_{s,B}=\frac12(100)(0.10)^2=0.5\,\text{J}
\]
Hence
\[
K_B=12.5-2-0.5=10\,\text{J}
\]
Therefore
\[
\frac12v_B^2=10,
\]
so
\[
v_B=2\sqrt5\,\text{m s}^{-1}
\]
The work-based check gives
\[
W_g=4\,\text{J},
\qquad
W_s=4\,\text{J},
\]
and therefore
\[
K_B=2+4+4=10\,\text{J}
\]
\(\textbf{Final answer:}\)
\[
\text{P, Q and S};\qquad v_B=2\sqrt5\,\text{m s}^{-1}.
\]
319. A force is known to be conservative. Its work in taking a particle from point A to point B along one allowed path is \(+18\,\text{J}\). Which set must be correct?
ⓐ. Work from A to B along another allowed path is \(+18\,\text{J}\), work from B to A is \(-18\,\text{J}\), and work over A \(\to\) B \(\to\) A is zero
ⓑ. Work from A to B along another path may differ, work from B to A is \(+18\,\text{J}\), and closed-path work is \(36\,\text{J}\)
ⓒ. Work from A to B along every path is zero because the force is conservative
ⓓ. Work from B to A is zero, but work over a closed path may be non-zero
Correct Answer: Work from A to B along another allowed path is \(+18\,\text{J}\), work from B to A is \(-18\,\text{J}\), and work over A \(\to\) B \(\to\) A is zero
Explanation: For a conservative force, work between two fixed endpoints is path independent.
Therefore every allowed A-to-B path gives the same work:
\[
W_{A\to B}=+18\,\text{J}
\]
Reversing the endpoints reverses the sign of the work:
\[
W_{B\to A}=-18\,\text{J}
\]
Hence the work around the closed route A \(\to\) B \(\to\) A is
\[
W_{\text{closed}}=18-18=0
\]
Zero closed-path work does not mean that the force performs zero work on every individual segment.
It means the positive and negative conservative-work contributions sum to zero over a complete return to the starting configuration.
320. Consider the following statements about conservative forces.
Statement I: The work done by a conservative force between two fixed endpoints depends only on those endpoints and not on the particular allowed path.
Statement II: The work done by a conservative force around a closed path is zero.
Statement III: If a conservative force has zero work around a closed path, its work must be zero on every segment of that path.
Statement IV: Different allowed paths joining the same two endpoints can give different works for a conservative force.
Which of the following is correct?
ⓐ. Statements I and II only
ⓑ. Statements I, II and III only
ⓒ. Statements III and IV only
ⓓ. Statements I and IV only
Correct Answer: Statements I and II only
Explanation: Statement I is the defining path-independence property of a conservative force.
For such a force,
\[
W_{A\to B}
\]
is fixed once the initial and final configurations are fixed.
Statement II is an equivalent consequence: returning to the starting point gives zero total conservative-force work.
Statement III is incorrect because individual parts of the closed route may involve positive or negative work.
Only their algebraic sum over the complete loop is required to vanish.
Statement IV is also incorrect because it directly contradicts path independence.
Near-Earth gravity and the ideal spring force are standard examples of conservative forces in this Unit.
