1. A constant force of magnitude \(20\,\text{N}\) acts on a body whose displacement is \(3\,\text{m}\). The angle between the force and displacement is \(60^\circ\). The work done by the force is
ⓐ. \(60\,\text{J}\)
ⓑ. \(30\,\text{J}\)
ⓒ. \(20\sqrt3\,\text{J}\)
ⓓ. \(0\,\text{J}\)
Correct Answer: \(30\,\text{J}\)
Explanation: Mechanical work by a constant force is the scalar product of force and displacement.
\(\textbf{Governing relation:}\)
\[
W=\vec F\cdot\vec s=Fs\cos\theta
\]
Only the component \(F\cos\theta\) parallel to the displacement contributes to work.
\(\textbf{Substitution:}\)
\[
W=(20)(3)\cos60^\circ
\]
\[
\boxed{W=(20)(3)\left(\frac12\right)=30\,\text{J}}
\]
The perpendicular component of the force contributes no work.
The result is a scalar and therefore has no direction associated with it.
2. Consider the following statements about mechanical work.
Statement I: Work is a scalar quantity even though force and displacement are vector quantities.
Statement II: The SI unit of work can be written as \(\text{kg m s}^{-2}\).
ⓐ. Both Statement I and Statement II are correct
ⓑ. Both Statement I and Statement II are incorrect
ⓒ. Statement I is correct but Statement II is incorrect
ⓓ. Statement I is incorrect but Statement II is correct
Correct Answer: Statement I is correct but Statement II is incorrect
Explanation: Statement I is correct because mechanical work is defined through the scalar product \(\vec F\cdot\vec s\).
A scalar product produces a scalar even when both quantities entering it are vectors.
Thus work has magnitude and algebraic sign but no spatial direction.
Statement II is incorrect because \(\text{kg m s}^{-2}\) is the SI unit of force, namely the newton.
For work, force must additionally be multiplied by displacement.
Hence
\[
1\,\text{J}=1\,\text{N m}=1\,\text{kg m}^2\text{s}^{-2}.
\]
The dimensions of work are therefore \([ML^2T^{-2}]\).
Thus only Statement I is correct.
3. A constant force \(\vec F=(3\hat i+4\hat j)\,\text{N}\) produces a displacement \(\vec s=(4\hat i-3\hat j)\,\text{m}\). The work done by the force is
ⓐ. \(25\,\text{J}\)
ⓑ. \(7\,\text{J}\)
ⓒ. \(0\,\text{J}\)
ⓓ. \(-25\,\text{J}\)
Correct Answer: \(0\,\text{J}\)
Explanation: For vectors specified in component form, work is obtained from their scalar product.
\(\textbf{Component form:}\)
\[
W=\vec F\cdot\vec s=F_xs_x+F_ys_y
\]
\(\textbf{Substitution:}\)
\[
W=(3)(4)+(4)(-3)
\]
\[
\boxed{W=12-12=0\,\text{J}}
\]
Both the force and displacement are non-zero, yet their scalar product vanishes.
Indeed, the two vectors are perpendicular because \(\vec F\cdot\vec s=0\).
This illustrates that a non-zero force does not necessarily perform non-zero work.
It is the component of force along the displacement, not merely the force magnitude, that determines work.
4. A force of magnitude \(F\) produces displacement \(s\) while making an angle \(60^\circ\) with the displacement. In a second case, the displacement remains \(s\), the force magnitude becomes \(2F\), and the angle becomes \(120^\circ\). If the works in the two cases are \(W_1\) and \(W_2\), respectively, then
ⓐ. \(W_2=2W_1\)
ⓑ. \(W_2=W_1\)
ⓒ. \(W_2=-W_1\)
ⓓ. \(W_2=-2W_1\)
Correct Answer: \(W_2=-2W_1\)
Explanation: Work depends on both force magnitude and the cosine of the angle between force and displacement.
\(\textbf{First case:}\)
\[
W_1=Fs\cos60^\circ=\frac{Fs}{2}
\]
The first work is positive because the force has a component along the displacement.
\(\textbf{Second case:}\)
\[
W_2=(2F)s\cos120^\circ
\]
Since \(\cos120^\circ=-\frac12\),
\[
W_2=(2F)s\left(-\frac12\right)=-Fs
\]
Using \(W_1=Fs/2\),
\[
\boxed{W_2=-2W_1}
\]
Doubling the force does not simply double the original positive work because the changed angle also reverses the sign of the parallel component.
5. Assertion: In uniform circular motion, the centripetal force can do zero work at every instant even though the velocity of the particle changes continuously.
Reason: The centripetal force is perpendicular to the instantaneous displacement \(d\vec r\), so \(\vec F\cdot d\vec r=0\).
ⓐ. Both Assertion and Reason are correct, and the Reason correctly explains the Assertion
ⓑ. Both Assertion and Reason are correct, but the Reason does not correctly explain the Assertion
ⓒ. Assertion is correct, but Reason is incorrect
ⓓ. Assertion is incorrect, but Reason is correct
Correct Answer: Both Assertion and Reason are correct, and the Reason correctly explains the Assertion
Explanation: The Assertion is correct because a force can alter the direction of velocity without changing the speed.
In uniform circular motion, the instantaneous velocity is tangential to the circular path.
The centripetal force is directed radially inward.
The instantaneous displacement \(d\vec r\) is along the tangent and is therefore perpendicular to the centripetal force.
Hence
\[
dW=\vec F\cdot d\vec r=F\,dr\cos90^\circ=0.
\]
Thus the centripetal force transfers no energy through mechanical work.
Nevertheless, it changes the direction of the velocity vector continuously.
Zero work therefore does not imply that a force has no dynamical effect.
The Reason is true and gives the direct physical explanation of the Assertion.
6. Consider the following statements.
Statement I: If the displacement of a body is zero during an interval, the work done on it by every finite force during that interval is zero.
Statement II: If the net work on a body is zero, every individual force acting on it must have done zero work.
Statement III: Negative work does not specify a spatial direction; it indicates that the selected force has a component opposite to the displacement.
Statement IV: A force may do zero work while continuously changing the direction of the velocity.
Which of the following is correct?
ⓐ. Statements I and II only
ⓑ. Statements I, III and IV only
ⓒ. Statements II, III and IV only
ⓓ. Statements I, II, III and IV
Correct Answer: Statements I, III and IV only
Explanation: Statement I is correct because work contains the displacement factor, so zero displacement gives zero mechanical work for every finite force.
Statement II is incorrect because different forces can perform positive and negative work that cancel algebraically.
For example, one force may do \(+W\) while another does \(-W\), giving zero net work even though neither individual work is zero.
Statement III is correct because work is a scalar quantity.
Its negative sign indicates the relative orientation of the force component and displacement, not a vector direction of work.
Statement IV is also correct.
A force perpendicular to the instantaneous motion can have zero power and zero work while changing the direction of velocity.
Uniform circular motion provides a standard example.
Therefore Statements I, III and IV are the correct combination.
7. A \(5\,\text{kg}\) block moves \(6\,\text{m}\) horizontally to the right. It is pulled by a constant \(50\,\text{N}\) force directed \(37^\circ\) above the horizontal, while a constant frictional force of \(20\,\text{N}\) opposes the motion. Take \(\cos37^\circ=4/5\). Which set correctly gives the work done by the pulling force, friction, gravity, normal reaction and all forces together?
ⓐ. \(300\,\text{J},-120\,\text{J},0,0,180\,\text{J}\)
ⓑ. \(240\,\text{J},+120\,\text{J},0,0,360\,\text{J}\)
ⓒ. \(240\,\text{J},-120\,\text{J},0,0,120\,\text{J}\)
ⓓ. \(240\,\text{J},-120\,\text{J},-300\,\text{J},+300\,\text{J},120\,\text{J}\)
Correct Answer: \(240\,\text{J},-120\,\text{J},0,0,120\,\text{J}\)
Explanation: \(\textbf{Pulling force:}\)
Only the horizontal component of the \(50\,\text{N}\) force contributes to work because the displacement is horizontal.
\[
W_P=Fs\cos37^\circ
\]
\[
W_P=(50)(6)\left(\frac45\right)=240\,\text{J}
\]
\(\textbf{Friction:}\)
Friction is opposite to the displacement, so its work is negative.
\[
W_f=-(20)(6)=-120\,\text{J}
\]
\(\textbf{Vertical forces:}\)
Gravity and the normal reaction are vertical while the displacement is horizontal.
\[
W_g=0,\qquad W_N=0
\]
Their magnitudes may be non-zero, but their scalar products with the displacement vanish.
\(\textbf{Net work:}\)
\[
W_{\text{net}}=240-120=120\,\text{J}
\]
Hence the correct set is \(240\,\text{J},-120\,\text{J},0,0,120\,\text{J}\).
8. A \(2\,\text{kg}\) block is pulled \(5\,\text{m}\) up a straight rough incline by a constant \(30\,\text{N}\) force parallel to the incline. During this displacement the block rises vertically by \(3\,\text{m}\), and a constant frictional force of \(4\,\text{N}\) acts down the incline. Take \(g=10\,\text{m s}^{-2}\). The net work done on the block is
ⓐ. \(210\,\text{J}\)
ⓑ. \(130\,\text{J}\)
ⓒ. \(90\,\text{J}\)
ⓓ. \(70\,\text{J}\)
Correct Answer: \(70\,\text{J}\)
Explanation: Several forces act, so the net work must be obtained by adding the work of each force algebraically.
\(\textbf{Applied force:}\)
The pulling force is parallel to the displacement.
\[
W_P=(30)(5)=150\,\text{J}
\]
\(\textbf{Gravity:}\)
For gravity, only the vertical change in position matters.
\[
W_g=-mgh
\]
\[
W_g=-(2)(10)(3)=-60\,\text{J}
\]
\(\textbf{Friction:}\)
Friction opposes the motion along the entire \(5\,\text{m}\) path.
\[
W_f=-(4)(5)=-20\,\text{J}
\]
The normal reaction is perpendicular to the displacement, so \(W_N=0\).
\(\textbf{Net work:}\)
\[
\boxed{W_{\text{net}}=150-60-20=70\,\text{J}}
\]
Using \(5\,\text{m}\) instead of the \(3\,\text{m}\) vertical rise in the gravitational-work calculation would incorrectly treat path length as height.
9. A block moves \(5\,\text{m}\) horizontally on a rough surface. A force of magnitude \(25\,\text{N}\) acts at an angle \(\theta\) above the direction of motion, while the kinetic-friction magnitude remains \(10\,\text{N}\). The net work done by these two forces is \(25\,\text{J}\). The force is then changed to act at \(90^\circ\) to the displacement, with its magnitude and the friction magnitude unchanged. Which pair gives \(\theta\) approximately and the new net work?
ⓐ. \(37^\circ,\,-50\,\text{J}\)
ⓑ. \(53^\circ,\,0\,\text{J}\)
ⓒ. \(53^\circ,\,+50\,\text{J}\)
ⓓ. \(53^\circ,\,-50\,\text{J}\)
Correct Answer: \(53^\circ,\,-50\,\text{J}\)
Explanation: \(\textbf{Initial work balance:}\)
The applied force contributes \(25\cos\theta\) along the displacement, while friction contributes \(-10\,\text{N}\).
\[
W_{\text{net}}=\left(25\cos\theta-10\right)(5)
\]
The initial net work is \(25\,\text{J}\), so
\[
\left(25\cos\theta-10\right)(5)=25
\]
Dividing by \(5\),
\[
25\cos\theta-10=5
\]
\[
25\cos\theta=15,
\qquad
\cos\theta=0.6
\]
Thus
\[
\theta\simeq53^\circ
\]
\(\textbf{Changed direction:}\)
At \(90^\circ\), the applied force does no work because \(\cos90^\circ=0\).
Friction still does
\[
W_f=-(10)(5)=-50\,\text{J}
\]
Therefore the new net work is \(-50\,\text{J}\), giving the pair \(53^\circ,-50\,\text{J}\).
10. A particle moves successively from \(O(0,0)\) to \(P(3,0)\), then to \(Q(3,4)\), and finally to \(R(1,4)\), with all coordinates in metres. A constant force \(\vec F=(6\hat i-2\hat j)\,\text{N}\) acts throughout. Which option gives the works done over \(OP\), \(PQ\), \(QR\), and the complete motion \(O\to R\), respectively?
ⓐ. \(18\,\text{J},\,8\,\text{J},\,-12\,\text{J},\,14\,\text{J}\)
ⓑ. \(18\,\text{J},\,-8\,\text{J},\,12\,\text{J},\,22\,\text{J}\)
ⓒ. \(18\,\text{J},\,-8\,\text{J},\,-12\,\text{J},\,-2\,\text{J}\)
ⓓ. \(18\,\text{J},\,-8\,\text{J},\,-12\,\text{J},\,-18\,\text{J}\)
Correct Answer: \(18\,\text{J},\,-8\,\text{J},\,-12\,\text{J},\,-2\,\text{J}\)
Explanation: Work by the constant vector force on each segment is \(\vec F\cdot\Delta\vec r\).
\(\textbf{Segment }OP\textbf{:}\)
\[
\Delta\vec r_{OP}=3\hat i
\]
\[
W_{OP}=(6\hat i-2\hat j)\cdot(3\hat i)=18\,\text{J}
\]
\(\textbf{Segment }PQ\textbf{:}\)
\[
\Delta\vec r_{PQ}=4\hat j
\]
\[
W_{PQ}=(6\hat i-2\hat j)\cdot(4\hat j)=-8\,\text{J}
\]
\(\textbf{Segment }QR\textbf{:}\)
\[
\Delta\vec r_{QR}=-2\hat i
\]
\[
W_{QR}=(6\hat i-2\hat j)\cdot(-2\hat i)=-12\,\text{J}
\]
Adding the segment works,
\[
W_{\text{total}}=18-8-12=-2\,\text{J}
\]
As a check, the net displacement is \(\hat i+4\hat j\), whose dot product with \(\vec F\) also gives \(6-8=-2\,\text{J}\).
11. During a short interval of contact between two bodies \(A\) and \(B\), the force exerted by \(B\) on \(A\) is \(30\,\text{N}\) toward the left. The force exerted by \(A\) on \(B\) is therefore \(30\,\text{N}\) toward the right. During the same interval, \(A\) moves \(2\,\text{m}\) to the right and \(B\) moves \(5\,\text{m}\) to the right. Which statement is correct?
ⓐ. The total work done by the two mutual forces is \(+210\,\text{J}\)
ⓑ. The total work done by the two mutual forces is zero because the forces form a Newton's third-law pair
ⓒ. The total work done by the two mutual forces is \(-210\,\text{J}\)
ⓓ. The total work done by the two mutual forces is \(+90\,\text{J}\)
Correct Answer: The total work done by the two mutual forces is \(+90\,\text{J}\)
Explanation: Newton's third law guarantees that the two mutual forces are equal in magnitude and opposite in direction.
It does not guarantee that the displacements of the two bodies are equal or opposite.
\(\textbf{Work on }A\textbf{:}\)
The force on \(A\) is leftward while its displacement is rightward.
\[
W_{B\to A}=-(30)(2)=-60\,\text{J}
\]
\(\textbf{Work on }B\textbf{:}\)
The force on \(B\) and its displacement are both rightward.
\[
W_{A\to B}=(30)(5)=150\,\text{J}
\]
\(\textbf{Total mutual work:}\)
\[
\boxed{W_{\text{mutual}}=-60+150=90\,\text{J}}
\]
The cancellation of forces in a third-law pair is a statement about force vectors, not automatically about their works.
Work depends separately on the displacement of the body on which each force acts.
12. A \(4\,\text{kg}\) block remains in contact with the horizontal floor of an elevator accelerating vertically upward at \(2\,\text{m s}^{-2}\). During the interval considered, the block moves upward through \(3\,\text{m}\). Take \(g=10\,\text{m s}^{-2}\). Which set gives the work done by the normal reaction, work done by gravity, and net work on the block?
ⓐ. \(120\,\text{J},-120\,\text{J},0\)
ⓑ. \(144\,\text{J},+120\,\text{J},264\,\text{J}\)
ⓒ. \(48\,\text{J},-40\,\text{J},8\,\text{J}\)
ⓓ. \(144\,\text{J},-120\,\text{J},24\,\text{J}\)
Correct Answer: \(144\,\text{J},-120\,\text{J},24\,\text{J}\)
Explanation: This situation is important because the normal reaction is not perpendicular to the block's displacement.
\(\textbf{Find the normal force:}\)
For upward acceleration,
\[
N-mg=ma
\]
Therefore
\[
N=m(g+a)=4(10+2)=48\,\text{N}
\]
\(\textbf{Work by the normal reaction:}\)
The normal reaction and displacement are both upward.
\[
W_N=Ns=(48)(3)=144\,\text{J}
\]
\(\textbf{Work by gravity:}\)
Gravity is opposite to the upward displacement.
\[
W_g=-mgs=-(4)(10)(3)=-120\,\text{J}
\]
\(\textbf{Net work:}\)
\[
W_{\text{net}}=144-120=24\,\text{J}
\]
The result also agrees with \(F_{\text{net}}s=(ma)s=(4)(2)(3)=24\,\text{J}\).
Thus a normal reaction does not universally perform zero work; its work depends on its orientation relative to the displacement.
13. A \(2\,\text{kg}\) pendulum bob is attached to a light string of length \(1.5\,\text{m}\). It moves from the lowest point to a position where the string makes \(60^\circ\) with the downward vertical. Take \(g=10\,\text{m s}^{-2}\). The work done by tension and by gravity during this motion are, respectively,
ⓐ. \(0,\,+15\,\text{J}\)
ⓑ. \(0,\,-15\,\text{J}\)
ⓒ. \(+15\,\text{J},\,-15\,\text{J}\)
ⓓ. \(-15\,\text{J},\,0\)
Correct Answer: \(0,\,-15\,\text{J}\)
Explanation: \(\textbf{Work by tension:}\)
At every point of the circular arc, tension acts radially along the string.
The instantaneous displacement of the bob is tangential to the arc.
Therefore tension is perpendicular to the instantaneous displacement throughout the motion.
\[
W_T=0
\]
\(\textbf{Vertical rise of the bob:}\)
At angle \(\theta\) from the downward vertical, the rise above the lowest point is
\[
h=L(1-\cos\theta)
\]
For \(L=1.5\,\text{m}\) and \(\theta=60^\circ\),
\[
h=1.5\left(1-\frac12\right)=0.75\,\text{m}
\]
\(\textbf{Work by gravity:}\)
\[
W_g=-mgh
\]
\[
W_g=-(2)(10)(0.75)=-15\,\text{J}
\]
The negative sign follows because the bob rises while gravity acts downward.
14. A body undergoes a displacement of \(4\,\text{m}\) along the positive \(x\)-direction while the following constant forces act on it.
| Force | Magnitude | Angle with \(+x\) |
|---|
| P | \(10\,\text{N}\) | \(0^\circ\) |
| Q | \(20\,\text{N}\) | \(60^\circ\) |
| R | \(6\,\text{N}\) | \(180^\circ\) |
| S | \(12\,\text{N}\) | \(90^\circ\) |
Which option gives \((W_P,W_Q,W_R,W_S,W_{\text{net}})\)?
ⓐ. \((40,80,-24,0,96)\,\text{J}\)
ⓑ. \((40,40,+24,0,104)\,\text{J}\)
ⓒ. \((40,40,-24,0,56)\,\text{J}\)
ⓓ. \((40,40,-24,48,104)\,\text{J}\)
Correct Answer: \((40,40,-24,0,56)\,\text{J}\)
Explanation: For every force, only its component parallel to the \(+x\) displacement contributes to work.
\(\textbf{Force P:}\)
\[
W_P=(10)(4)\cos0^\circ=40\,\text{J}
\]
\(\textbf{Force Q:}\)
\[
W_Q=(20)(4)\cos60^\circ=40\,\text{J}
\]
\(\textbf{Force R:}\)
\[
W_R=(6)(4)\cos180^\circ=-24\,\text{J}
\]
\(\textbf{Force S:}\)
\[
W_S=(12)(4)\cos90^\circ=0
\]
Force \(S\) is non-zero but has no component along the displacement.
\(\textbf{Net work:}\)
\[
W_{\text{net}}=40+40-24+0=56\,\text{J}
\]
The signs of the individual works must be retained when they are combined.
15. A constant horizontal force of \(12\,\text{N}\) acts eastward on a particle. The particle first moves along the upper semicircle of a circle of radius \(2\,\text{m}\), from its westernmost point to its easternmost point. It then returns to the starting point along the lower semicircle while the same constant force continues to act. The work done by this force during the first semicircle and during the complete round trip are, respectively,
ⓐ. \(24\pi\,\text{J},\,0\)
ⓑ. \(48\,\text{J},\,48\,\text{J}\)
ⓒ. \(0,\,0\)
ⓓ. \(48\,\text{J},\,0\)
Correct Answer: \(48\,\text{J},\,0\)
Explanation: For a constant vector force, the work over any path can be evaluated from the net displacement.
\(\textbf{First semicircle:}\)
The diameter is
\[
2R=4\,\text{m}
\]
The particle's net displacement from the westernmost to the easternmost point is \(4\,\text{m}\) eastward.
The applied force is also eastward, so
\[
W_1=F\Delta x=(12)(4)=48\,\text{J}
\]
The semicircular distance \(2\pi\,\text{m}\) must not be substituted in place of displacement for this constant horizontal force.
\(\textbf{Return motion:}\)
On the lower semicircle, the net displacement is \(4\,\text{m}\) westward.
Hence
\[
W_2=(12)(-4)=-48\,\text{J}
\]
\(\textbf{Complete trip:}\)
\[
W_{\text{round}}=48-48=0
\]
The particle returns to its starting point, so its total displacement is zero and the total work by this constant force is zero.
16. A particle is constrained to move once around a rectangular path of sides \(4\,\text{m}\) and \(3\,\text{m}\), returning to its starting point. A constant external force of \(10\,\text{N}\) acts eastward throughout. A kinetic-friction force of constant magnitude \(2\,\text{N}\) always acts opposite to the instantaneous motion. Any constraining forces do no work. Which set gives the work by the constant eastward force, the work by friction, and the net work over the complete circuit?
ⓐ. \(0,\,+28\,\text{J},\,+28\,\text{J}\)
ⓑ. \(140\,\text{J},-28\,\text{J},112\,\text{J}\)
ⓒ. \(0,-14\,\text{J},-14\,\text{J}\)
ⓓ. \(0,-28\,\text{J},-28\,\text{J}\)
Correct Answer: \(0,-28\,\text{J},-28\,\text{J}\)
Explanation: The two forces must be treated differently because the eastward force has a fixed direction whereas friction continually opposes the motion.
\(\textbf{Constant eastward force:}\)
After one complete circuit, the particle returns to its initial position.
Therefore its net displacement is zero.
For a constant vector force,
\[
W_F=\vec F\cdot\Delta\vec r=0
\]
Using the perimeter as the displacement for this force would be incorrect.
\(\textbf{Frictional work:}\)
Friction is opposite to the motion on every segment, so its work is negative throughout.
The total distance travelled is the rectangle's perimeter:
\[
s_{\text{total}}=2(4+3)=14\,\text{m}
\]
Hence
\[
W_f=-fs_{\text{total}}=-(2)(14)=-28\,\text{J}
\]
\(\textbf{Net work:}\)
\[
W_{\text{net}}=W_F+W_f=0-28=-28\,\text{J}
\]
Thus returning to the starting point makes the work of the fixed constant force zero, but it does not make the work of friction zero.
17. A constant force of magnitude \(40\,\text{N}\) does \(100\,\text{J}\) of work while a particle undergoes a displacement of \(5\,\text{m}\). The angle between the force and this displacement is \(\theta\). The force is then kept unchanged while the particle undergoes a \(2\,\text{m}\) displacement exactly opposite to its original displacement direction. Which pair gives \(\theta\) and the work done by the force during the second displacement?
ⓐ. \(60^\circ,\,+40\,\text{J}\)
ⓑ. \(120^\circ,\,-40\,\text{J}\)
ⓒ. \(60^\circ,\,-40\,\text{J}\)
ⓓ. \(60^\circ,\,-80\,\text{J}\)
Correct Answer: \(60^\circ,\,-40\,\text{J}\)
Explanation: \(\textbf{Infer the original angle:}\)
For the first displacement,
\[
W_1=Fs_1\cos\theta
\]
Substituting the given values,
\[
100=(40)(5)\cos\theta
\]
\[
\cos\theta=\frac{100}{200}=\frac12
\]
Hence
\[
\theta=60^\circ
\]
\(\textbf{Reverse the displacement:}\)
The force direction is unchanged, but the second displacement is exactly opposite to the original displacement.
Therefore the new angle between the force and displacement is
\[
180^\circ-60^\circ=120^\circ
\]
Thus
\[
W_2=(40)(2)\cos120^\circ
\]
\[
W_2=(80)\left(-\frac12\right)=-40\,\text{J}
\]
The reversal of displacement changes the sign of the projection of the unchanged force along the motion.
\(\textbf{Final answer:}\)
\[
\boxed{\theta=60^\circ,\qquad W_2=-40\,\text{J}}
\]
18. Two masses \(2\,\text{kg}\) and \(6\,\text{kg}\) are connected by a light inextensible string passing over an ideal pulley. During a certain interval, the \(2\,\text{kg}\) mass rises \(3\,\text{m}\) while the \(6\,\text{kg}\) mass descends \(3\,\text{m}\). The string tension is \(30\,\text{N}\) during this motion. Take \(g=10\,\text{m s}^{-2}\). For the two-mass system, the total work done by tension and the total work done by gravity are, respectively,
ⓐ. \(-180\,\text{J},\,+120\,\text{J}\)
ⓑ. \(0,\,-120\,\text{J}\)
ⓒ. \(+180\,\text{J},\,+120\,\text{J}\)
ⓓ. \(0,\,+120\,\text{J}\)
Correct Answer: \(0,\,+120\,\text{J}\)
Explanation: The work of each force must first be associated with the body on which that force acts.
\(\textbf{Tension on the }2\,\text{kg}\textbf{ mass:}\)
Tension and displacement are both upward.
\[
W_{T,1}=Ts=(30)(3)=+90\,\text{J}
\]
\(\textbf{Tension on the }6\,\text{kg}\textbf{ mass:}\)
Tension is upward while this mass moves downward.
\[
W_{T,2}=-Ts=-(30)(3)=-90\,\text{J}
\]
Hence, for the complete two-mass system,
\[
W_T=+90-90=0
\]
Tension does non-zero work on each mass separately, even though its total work on this particular system is zero.
\(\textbf{Gravity on the rising mass:}\)
\[
W_{g,1}=-(2)(10)(3)=-60\,\text{J}
\]
\(\textbf{Gravity on the descending mass:}\)
\[
W_{g,2}=+(6)(10)(3)=+180\,\text{J}
\]
Therefore
\[
W_g=-60+180=+120\,\text{J}
\]
\(\textbf{Final answer:}\)
\[
\boxed{W_T=0,\qquad W_g=+120\,\text{J}}
\]
19. A block first moves \(6\,\text{m}\) to the right and then \(2\,\text{m}\) to the left along the same horizontal line. A constant applied force of \(25\,\text{N}\) acts to the right throughout. A kinetic-friction force of constant magnitude \(5\,\text{N}\) always opposes the instantaneous motion. Which option gives the total work done by the applied force, the total work done by friction, and their combined work?
ⓐ. \(200\,\text{J},-40\,\text{J},160\,\text{J}\)
ⓑ. \(100\,\text{J},-20\,\text{J},80\,\text{J}\)
ⓒ. \(200\,\text{J},-20\,\text{J},180\,\text{J}\)
ⓓ. \(100\,\text{J},-40\,\text{J},60\,\text{J}\)
Correct Answer: \(100\,\text{J},-40\,\text{J},60\,\text{J}\)
Explanation: The applied force has a fixed direction, whereas the frictional force reverses direction when the motion reverses.
\(\textbf{Applied force over the first segment:}\)
\[
W_{F,1}=(25)(6)=150\,\text{J}
\]
\(\textbf{Applied force over the return segment:}\)
The displacement is now opposite to the applied force.
\[
W_{F,2}=-(25)(2)=-50\,\text{J}
\]
Thus
\[
W_F=150-50=100\,\text{J}
\]
This is also \(25\,\text{N}\) multiplied by the net rightward displacement of \(4\,\text{m}\).
\(\textbf{Frictional work:}\)
Friction opposes motion on both segments, so both contributions are negative.
\[
W_f=-(5)(6)-(5)(2)
\]
\[
W_f=-30-10=-40\,\text{J}
\]
For friction here, the relevant accumulated path length is \(6+2=8\,\text{m}\), not the net displacement.
\(\textbf{Combined work:}\)
\[
W_F+W_f=100-40=60\,\text{J}
\]
\(\textbf{Final answer:}\)
\[
\boxed{100\,\text{J},-40\,\text{J},60\,\text{J}}
\]
20. A particle undergoes a \(10\,\text{m}\) displacement along the positive \(x\)-direction. Three constant forces act on it: \(20\,\text{N}\) at \(60^\circ\) to the displacement, \(15\,\text{N}\) exactly opposite to the displacement, and \(10\,\text{N}\) perpendicular to the displacement. A fourth constant force of magnitude \(10\,\text{N}\) makes an angle \(\theta\) with the displacement. If the total work of all four forces is zero, and the fourth force is then reversed while all other conditions remain unchanged, which pair gives \(\theta\) and the new total work?
ⓐ. \(30^\circ,\,-50\,\text{J}\)
ⓑ. \(60^\circ,\,-100\,\text{J}\)
ⓒ. \(90^\circ,\,-50\,\text{J}\)
ⓓ. \(120^\circ,\,-100\,\text{J}\)
Correct Answer: \(60^\circ,\,-100\,\text{J}\)
Explanation: \(\textbf{Work of the first three forces:}\)
For the \(20\,\text{N}\) force,
\[
W_1=(20)(10)\cos60^\circ=100\,\text{J}
\]
For the \(15\,\text{N}\) opposing force,
\[
W_2=(15)(10)\cos180^\circ=-150\,\text{J}
\]
The perpendicular force does no work:
\[
W_3=(10)(10)\cos90^\circ=0
\]
Hence the first three forces together do
\[
W_{123}=100-150=-50\,\text{J}
\]
\(\textbf{Condition for zero total work:}\)
The fourth force must therefore do \(+50\,\text{J}\).
\[
(10)(10)\cos\theta=50
\]
\[
\cos\theta=\frac12,
\qquad
\theta=60^\circ
\]
\(\textbf{Reverse the fourth force:}\)
Reversing this force changes its work from \(+50\,\text{J}\) to \(-50\,\text{J}\).
The contribution of the other three forces remains \(-50\,\text{J}\).
Therefore
\[
W'_{\text{total}}=-50-50=-100\,\text{J}
\]
\(\textbf{Final answer:}\)
\[
\boxed{\theta=60^\circ,\qquad W'_{\text{total}}=-100\,\text{J}}
\]
