201. A \(4\,\text{kg}\) block moves \(4\,\text{m}\) horizontally with initial speed \(2\,\text{m s}^{-1}\). A \(25\,\text{N}\) force pulls it at \(37^\circ\) above the horizontal. The block's final speed is \(2\sqrt6\,\text{m s}^{-1}\). Take \(g=10\,\text{m s}^{-2}\), \(\sin37^\circ=0.6\), and \(\cos37^\circ=0.8\). If the coefficient of kinetic friction is \(\mu\), then in a second trial the same \(25\,\text{N}\) force acts horizontally while all other initial conditions remain unchanged. What are \(\mu\) and the final speed in the second trial?
ⓐ. \(\mu=0.4,\;\sqrt{22}\,\text{m s}^{-1}\)
ⓑ. \(\mu=0.25,\;2\sqrt6\,\text{m s}^{-1}\)
ⓒ. \(\mu=0.4,\;5\,\text{m s}^{-1}\)
ⓓ. \(\mu=0.5,\;\sqrt{18}\,\text{m s}^{-1}\)
Correct Answer: \(\mu=0.4,\;\sqrt{22}\,\text{m s}^{-1}\)
Explanation: \(\textbf{Infer friction from Trial I:}\)
The initial kinetic energy is
\[
K_i=\frac12(4)(2^2)=8\,\text{J}
\]
The stated final speed gives
\[
K_f=\frac12(4)(2\sqrt6)^2
\]
\[
K_f=48\,\text{J}
\]
Thus
\[
W_{\text{net}}=40\,\text{J}
\]
The pulling force does
\[
W_P=(25)(4)(0.8)=80\,\text{J}
\]
The normal reaction is
\[
N=mg-F\sin37^\circ
\]
\[
N=40-25(0.6)=25\,\text{N}
\]
Therefore the frictional work is
\[
W_f=-(\mu N)(4)=-100\mu
\]
Hence
\[
80-100\mu=40
\]
Thus
\[
\mu=0.4
\]
\(\textbf{Trial II — horizontal force:}\)
Now
\[
N=mg=40\,\text{N},
\]
so
\[
f_k=\mu N=(0.4)(40)=16\,\text{N}
\]
The applied-force work is
\[
W_P'=(25)(4)=100\,\text{J},
\]
and friction does
\[
W_f'=-(16)(4)=-64\,\text{J}
\]
Therefore
\[
W_{\text{net}}'=36\,\text{J}
\]
Thus
\[
K_f'=8+36=44\,\text{J}
\]
Since
\[
44=\frac12(4)v'^2=2v'^2,
\]
we get
\[
v'=\sqrt{22}\,\text{m s}^{-1}
\]
\(\textbf{Final answer:}\)
\[
\boxed{\mu=0.4,\qquad v'=\sqrt{22}\,\text{m s}^{-1}}
\]
202. A \(2\,\text{kg}\) particle moves along a straight line with initial speed \(2\,\text{m s}^{-1}\). A constant net force of \(6\,\text{N}\) acts along its motion while it travels \(4\,\text{m}\). Which pair gives its final speed and the time taken for the \(4\,\text{m}\) displacement?
ⓐ. \(2\sqrt7\,\text{m s}^{-1},\;\dfrac{2(\sqrt7-1)}{3}\,\text{s}\)
ⓑ. \(2\sqrt7\,\text{m s}^{-1},\;\dfrac43\,\text{s}\)
ⓒ. \(4\,\text{m s}^{-1},\;2\,\text{s}\)
ⓓ. \(2\sqrt5\,\text{m s}^{-1},\;\dfrac{2(\sqrt5-1)}{3}\,\text{s}\)
Correct Answer: \(2\sqrt7\,\text{m s}^{-1},\;\dfrac{2(\sqrt7-1)}{3}\,\text{s}\)
Explanation: Work-energy reasoning efficiently gives the endpoint speed.
\(\textbf{Use work-energy for the speed:}\)
\[
W_{\text{net}}=Fs=(6)(4)=24\,\text{J}
\]
The initial kinetic energy is
\[
K_i=\frac12(2)(2^2)=4\,\text{J}
\]
Thus
\[
K_f=4+24=28\,\text{J}
\]
Hence
\[
28=\frac12(2)v^2=v^2,
\]
so
\[
v=2\sqrt7\,\text{m s}^{-1}
\]
The work-energy theorem alone has not supplied the elapsed time.
Here, however, the additional information that the net force is constant gives a constant acceleration:
\[
a=\frac{F}{m}=\frac62=3\,\text{m s}^{-2}
\]
Now the kinematic relation
\[
v=u+at
\]
may be used:
\[
2\sqrt7=2+3t
\]
Therefore
\[
t=\frac{2(\sqrt7-1)}{3}\,\text{s}
\]
\(\textbf{Final answer:}\)
\[
\boxed{v=2\sqrt7\,\text{m s}^{-1},\qquad t=\frac{2(\sqrt7-1)}{3}\,\text{s}}
\]
203. Consider a particle moving continuously toward increasing \(x\) under a one-dimensional net force \(F_{\text{net}}(x)\).
Statement I: Where the force is positive, the kinetic energy increases with increasing \(x\).
Statement II: If \(F_{\text{net}}=0\) at a position, the kinetic energy must also be zero there.
Statement III: If \(F_{\text{net}}\) changes sign from positive to negative as \(x\) increases and the particle passes through that position, its kinetic energy has a local maximum there.
Statement IV: Any finite interval over which \(F_{\text{net}}\lt 0\) must necessarily contain a stopping point.
Which of the following is correct?
ⓐ. Statements II and IV only
ⓑ. Statements I, II and III only
ⓒ. Statements I and III only
ⓓ. Statements I, III and IV only
Correct Answer: Statements I and III only
Explanation: For one-dimensional motion toward increasing \(x\),
\[
dK=F_{\text{net}}\,dx.
\]
Thus a positive net force corresponds to increasing kinetic energy with position, so Statement I is correct.
Statement II is incorrect.
A zero net force at one position means the local rate of change of kinetic energy with position is zero there; it does not require
\[
K=0.
\]
Statement III is correct.
If the force changes from positive to negative, kinetic energy changes from increasing to decreasing, so the passed point is a local maximum of \(K\).
Statement IV is incorrect.
A negative-force interval removes kinetic energy, but the particle stops only if the accumulated negative work is large enough to reduce its kinetic energy to zero.
A particle entering with sufficiently large kinetic energy can cross the entire negative-force interval without stopping.
Hence only Statements I and III are correct.
204.
A \(2\,\text{kg}\) particle enters a smooth track with speed \(4\,\text{m s}^{-1}\) and moves to a point \(5\,\text{m}\) vertically below its starting point. The track may have any shape. The normal reaction is everywhere perpendicular to the instantaneous displacement. Take \(g=10\,\text{m s}^{-2}\).
In Trial I the track is smooth throughout. In Trial II the same endpoints are used, but an additional resistive force does \(-20\,\text{J}\) of work during the motion. The final speeds in Trials I and II are
ⓐ. \(2\sqrt{29}\,\text{m s}^{-1}\) and \(4\sqrt6\,\text{m s}^{-1}\)
ⓑ. \(4\sqrt6\,\text{m s}^{-1}\) and \(2\sqrt{29}\,\text{m s}^{-1}\)
ⓒ. \(2\sqrt{21}\,\text{m s}^{-1}\) and \(2\sqrt{19}\,\text{m s}^{-1}\)
ⓓ. \(2\sqrt{29}\,\text{m s}^{-1}\) and \(2\sqrt{21}\,\text{m s}^{-1}\)
Correct Answer: \(2\sqrt{29}\,\text{m s}^{-1}\) and \(4\sqrt6\,\text{m s}^{-1}\)
Explanation: The normal reaction does zero work because it is perpendicular to the instantaneous displacement.
The initial kinetic energy is
\[
K_i=\frac12(2)(4^2)=16\,\text{J}
\]
\(\textbf{Trial I — smooth track:}\)
The \(5\,\text{m}\) descent gives gravitational work
\[
W_g=mgh
\]
\[
W_g=(2)(10)(5)=100\,\text{J}
\]
Thus
\[
K_f=16+100=116\,\text{J}
\]
For \(m=2\,\text{kg}\),
\[
v_1^2=116
\]
Hence
\[
v_1=2\sqrt{29}\,\text{m s}^{-1}
\]
\(\textbf{Trial II — additional resistance:}\)
The gravitational work is still \(100\,\text{J}\), but the resistance contributes
\[
W_R=-20\,\text{J}
\]
Therefore
\[
K_f'=16+100-20=96\,\text{J}
\]
Thus
\[
v_2^2=96,
\]
giving
\[
v_2=4\sqrt6\,\text{m s}^{-1}
\]
The shape of the smooth track does not affect gravitational work between fixed heights, while the added resistance lowers the final kinetic energy.
\(\textbf{Final answer:}\)
\[
\boxed{2\sqrt{29}\,\text{m s}^{-1},\;4\sqrt6\,\text{m s}^{-1}}
\]
205. Assertion: If the net work done on a particle over a finite displacement interval is negative, its speed need not have decreased monotonically throughout that entire interval.
Reason: Different subintervals may contribute positive and negative net work, while the work-energy theorem constrains the final-minus-initial kinetic energy through only their algebraic sum.
ⓐ. 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.
Negative total net work means only that
\[
K_f\lt K_i
\]
for the complete stated interval.
It does not require kinetic energy to decrease at every intermediate stage.
For example, the net force may do positive work in an early part of the motion, temporarily increasing the particle's kinetic energy.
A later region may then do a larger amount of negative work.
The algebraic total can be negative even though the speed first increased and only later decreased.
The Reason states exactly this cumulative-work principle.
The work-energy theorem gives
\[
K_f-K_i=\sum W_{\text{subintervals}},
\]
so endpoint comparison alone does not specify monotonic behaviour between the endpoints.
Therefore both the Assertion and Reason are correct, and the Reason properly explains the Assertion.
206. A machine transfers \(3.6\times10^4\,\text{J}\) of mechanical energy in \(30\,\text{s}\). Its average power during this interval is
ⓐ. \(0.12\,\text{kW}\)
ⓑ. \(1.2\,\text{kW}\)
ⓒ. \(12\,\text{kW}\)
ⓓ. \(108\,\text{kW}\)
Correct Answer: \(1.2\,\text{kW}\)
Explanation: Average power measures the rate at which work is done or energy is transferred over a finite time interval.
\[
P_{\text{avg}}=\frac{W}{\Delta t}
\]
Substituting the data,
\[
P_{\text{avg}}=\frac{3.6\times10^4}{30}
\]
\[
P_{\text{avg}}=1.2\times10^3\,\text{W}
\]
Since
\[
1\,\text{kW}=10^3\,\text{W},
\]
we obtain
\[
P_{\text{avg}}=1.2\,\text{kW}
\]
Power is not the amount of energy transferred; it is the rate of that transfer.
\(\textbf{Final answer:}\)
\[
\boxed{1.2\,\text{kW}}
\]
207. Consider the following statements about mechanical power.
Statement I: Average power over a time interval is the total work done divided by the duration of that interval.
Statement II: If a machine has average power \(P\) over an interval, its instantaneous power must equal \(P\) at every instant of that interval.
ⓐ. Both Statement I and Statement II are correct
ⓑ. Statement I is correct but Statement II is incorrect
ⓒ. Statement I is incorrect but Statement II is correct
ⓓ. Both Statement I and Statement II are incorrect
Correct Answer: Statement I is correct but Statement II is incorrect
Explanation: Statement I is correct because average power is defined by
\[
P_{\text{avg}}=\frac{\Delta W}{\Delta t}.
\]
It compares the total work or energy transfer with the total elapsed time.
Instantaneous power, however, describes the rate at one particular instant:
\[
P=\frac{dW}{dt}.
\]
A machine may operate at a high rate during part of an interval and at a lower rate during another part.
Its average may still have one definite value even though the instantaneous value changes continuously.
Therefore knowing the average power does not imply that the machine operated at that same power at every instant.
Statement II is consequently incorrect.
Thus only Statement I is correct.
208. Consider the following statements.
I. \(1\,\text{W}=1\,\text{J s}^{-1}\).
II. The dimensions of power are \([ML^2T^{-3}]\).
III. Kilowatt-hour is a unit of power.
IV. \(1\,\text{hp}=746\,\text{W}\).
Which of the following is correct?
ⓐ. I and III only
ⓑ. II and IV only
ⓒ. I, II and IV only
ⓓ. I, II, III and IV
Correct Answer: I, II and IV only
Explanation: Statement I is correct because power is work done per unit time:
\[
1\,\text{W}=1\,\text{J s}^{-1}.
\]
For Statement II,
\[
[P]=\frac{[W]}{[t]}.
\]
Since
\[
[W]=[ML^2T^{-2}],
\]
we get
\[
[P]=[ML^2T^{-3}],
\]
so Statement II is correct.
Statement III is incorrect because multiplying power by time produces energy.
Thus kilowatt-hour is a unit of energy, not power.
Statement IV is the standard conversion
\[
1\,\text{hp}=746\,\text{W}.
\]
Hence Statements I, II and IV are correct.
209. Machines \(A\) and \(B\) each perform \(12\,\text{kJ}\) of work. Machine \(A\) takes \(20\,\text{s}\), while machine \(B\) takes \(30\,\text{s}\). Which statement is correct?
ⓐ. \(P_A:P_B=3:2\), while both machines transfer the same energy
ⓑ. \(P_A:P_B=2:3\), and \(A\) transfers more energy
ⓒ. \(P_A=P_B\), because the works are equal
ⓓ. \(P_A:P_B=3:2\), and \(A\) transfers \(3/2\) times as much energy
Correct Answer: \(P_A:P_B=3:2\), while both machines transfer the same energy
Explanation: Both machines perform the same work:
\[
W_A=W_B=12\,\text{kJ}.
\]
Their average powers are
\[
P_A=\frac{W}{20},
\qquad
P_B=\frac{W}{30}.
\]
Therefore
\[
\frac{P_A}{P_B}
=
\frac{W/20}{W/30}
=
\frac{30}{20}
=
\frac32
\]
Thus
\[
P_A:P_B=3:2
\]
Machine \(A\) is more powerful because it performs the same work in less time.
However, power and energy are different quantities.
The total energy transferred by the two machines is equal because their works are equal.
210. The work done by a device from \(t=0\) up to time \(t\) is
\[
W(t)=4t^2+6t,
\]
where \(W\) is in joules and \(t\) in seconds. Its average power over the first \(3\,\text{s}\) and its instantaneous power at \(t=3\,\text{s}\) are, respectively,
ⓐ. \(24\,\text{W},\;30\,\text{W}\)
ⓑ. \(30\,\text{W},\;18\,\text{W}\)
ⓒ. \(18\,\text{W},\;18\,\text{W}\)
ⓓ. \(18\,\text{W},\;30\,\text{W}\)
Correct Answer: \(18\,\text{W},\;30\,\text{W}\)
Explanation: Average and instantaneous power describe different aspects of the work rate.
\(\textbf{Work during the first }3\,\text{s}\textbf{:}\)
\[
W(3)=4(3)^2+6(3)
\]
\[
W(3)=36+18=54\,\text{J}
\]
Since \(W(0)=0\),
\[
P_{\text{avg}}=\frac{54}{3}=18\,\text{W}
\]
\(\textbf{Instantaneous rate at }t=3\,\text{s}\textbf{:}\)
\[
P(t)=\frac{dW}{dt}
\]
Therefore
\[
P(t)=8t+6
\]
At \(t=3\,\text{s}\),
\[
P(3)=8(3)+6=30\,\text{W}
\]
The larger instantaneous value shows that the rate of doing work has increased with time.
\(\textbf{Final answer:}\)
\[
\boxed{18\,\text{W},\;30\,\text{W}}
\]
211. A machine operates according to the following schedule.
| Time interval | Power |
|---|
| \(0\) to \(2\,\text{s}\) | \(200\,\text{W}\) |
| \(2\) to \(5\,\text{s}\) | \(400\,\text{W}\) |
| \(5\) to \(10\,\text{s}\) | \(160\,\text{W}\) |
The total energy transferred and the average power over the complete \(10\,\text{s}\) interval are
ⓐ. \(2000\,\text{J},\;200\,\text{W}\)
ⓑ. \(2400\,\text{J},\;240\,\text{W}\)
ⓒ. \(2400\,\text{J},\;400\,\text{W}\)
ⓓ. \(1600\,\text{J},\;160\,\text{W}\)
Correct Answer: \(2400\,\text{J},\;240\,\text{W}\)
Explanation: The energy transferred in each interval is
\[
W=P\Delta t
\]
\(\textbf{First interval:}\)
\[
W_1=(200)(2)=400\,\text{J}
\]
\(\textbf{Second interval:}\)
\[
W_2=(400)(3)=1200\,\text{J}
\]
\(\textbf{Third interval:}\)
\[
W_3=(160)(5)=800\,\text{J}
\]
Thus the total transferred energy is
\[
W_{\text{total}}=400+1200+800=2400\,\text{J}
\]
The average power must use total energy and total elapsed time:
\[
P_{\text{avg}}=\frac{2400}{10}=240\,\text{W}
\]
It is not obtained by simply selecting the largest or longest-lasting power value.
\(\textbf{Final answer:}\)
\[
\boxed{2400\,\text{J},\;240\,\text{W}}
\]
212. A machine does \(600\,\text{J}\) of work during its first \(3\,\text{s}\) and \(1400\,\text{J}\) during the next \(7\,\text{s}\). In a modified operation, the first stage is unchanged but the second \(1400\,\text{J}\) is completed in \(4\,\text{s}\). The ratio of the modified average power to the original average power is
ⓐ. \(5/7\)
ⓑ. \(7/10\)
ⓒ. \(10/7\)
ⓓ. \(7/5\)
Correct Answer: \(10/7\)
Explanation: The total work is unchanged between the two operations:
\[
W_{\text{total}}=600+1400=2000\,\text{J}
\]
\(\textbf{Original operation:}\)
The total time is
\[
t_1=3+7=10\,\text{s}
\]
Hence
\[
P_1=\frac{2000}{10}=200\,\text{W}
\]
\(\textbf{Modified operation:}\)
The total time becomes
\[
t_2=3+4=7\,\text{s}
\]
Thus
\[
P_2=\frac{2000}{7}\,\text{W}
\]
Therefore
\[
\frac{P_2}{P_1}
=
\frac{2000/7}{200}
=
\frac{10}{7}
\]
The total energy transfer is unchanged, but completing it in less time increases the average power.
\(\textbf{Final answer:}\)
\[
\boxed{\frac{P_2}{P_1}=\frac{10}{7}}
\]
213. A motor delivers a constant mechanical power of \(2.5\,\text{hp}\) for \(120\,\text{s}\). Using \(1\,\text{hp}=746\,\text{W}\), the mechanical work done is
ⓐ. \(89.52\,\text{kJ}\)
ⓑ. \(186.5\,\text{kJ}\)
ⓒ. \(223.8\,\text{kJ}\)
ⓓ. \(447.6\,\text{kJ}\)
Correct Answer: \(223.8\,\text{kJ}\)
Explanation: \(\textbf{Convert horsepower to watts:}\)
\[
P=(2.5)(746)
\]
\[
P=1865\,\text{W}
\]
For constant power,
\[
W=P\Delta t
\]
Therefore
\[
W=(1865)(120)
\]
\[
W=223800\,\text{J}
\]
Converting joules to kilojoules,
\[
W=223.8\,\text{kJ}
\]
Horsepower and watt are units of power, whereas the product of power and time gives energy or work.
\(\textbf{Final answer:}\)
\[
\boxed{223.8\,\text{kJ}}
\]
214. A \(1.5\,\text{kW}\) device operates at its stated power for \(40\,\text{min}\). The energy transferred is
ⓐ. \(40\,\text{kWh}=1.44\times10^8\,\text{J}\)
ⓑ. \(1.5\,\text{kWh}=5.4\times10^6\,\text{J}\)
ⓒ. \(0.6\,\text{kWh}=2.16\times10^6\,\text{J}\)
ⓓ. \(1\,\text{kWh}=3.6\times10^6\,\text{J}\)
Correct Answer: \(1\,\text{kWh}=3.6\times10^6\,\text{J}\)
Explanation: The operating time in hours is
\[
40\,\text{min}=\frac{40}{60}\,\text{h}=\frac23\,\text{h}
\]
Energy transferred is
\[
E=Pt
\]
Therefore
\[
E=(1.5\,\text{kW})\left(\frac23\,\text{h}\right)
\]
\[
E=1\,\text{kWh}
\]
Using the standard conversion,
\[
1\,\text{kWh}
=
(1000\,\text{W})(3600\,\text{s})
\]
Thus
\[
1\,\text{kWh}=3.6\times10^6\,\text{J}
\]
Kilowatt-hour is therefore an energy unit despite containing the word kilowatt.
\(\textbf{Final answer:}\)
\[
\boxed{1\,\text{kWh}=3.6\times10^6\,\text{J}}
\]
215. A \(3\,\text{hp}\) motor operates at constant mechanical power for \(10\,\text{min}\). Take \(1\,\text{hp}=746\,\text{W}\). Which pair correctly gives the energy transferred in kilowatt-hour and in joules?
ⓐ. \(0.373\,\text{kWh},\;1.3428\times10^6\,\text{J}\)
ⓑ. \(2.238\,\text{kWh},\;8.0568\times10^6\,\text{J}\)
ⓒ. \(0.2238\,\text{kWh},\;8.0568\times10^5\,\text{J}\)
ⓓ. \(0.746\,\text{kWh},\;2.6856\times10^6\,\text{J}\)
Correct Answer: \(0.373\,\text{kWh},\;1.3428\times10^6\,\text{J}\)
Explanation: \(\textbf{Convert the motor power:}\)
\[
P=(3)(746)=2238\,\text{W}
\]
Thus
\[
P=2.238\,\text{kW}
\]
The operating time is
\[
10\,\text{min}=\frac16\,\text{h}
\]
Therefore
\[
E=(2.238)\left(\frac16\right)\text{kWh}
\]
\[
E=0.373\,\text{kWh}
\]
Using
\[
1\,\text{kWh}=3.6\times10^6\,\text{J},
\]
we obtain
\[
E=(0.373)(3.6\times10^6)
\]
\[
E=1.3428\times10^6\,\text{J}
\]
The two numerical values represent the same energy expressed in different units.
\(\textbf{Final answer:}\)
\[
\boxed{0.373\,\text{kWh},\;1.3428\times10^6\,\text{J}}
\]
216. Two devices begin doing work at \(t=0\). Their cumulative works are
\[
W_A(t)=20t,\qquad W_B(t)=5t^2,
\]
where work is in joules and time in seconds. Over the interval \(0\leq t\leq4\,\text{s}\), which statement is correct?
ⓐ. Device \(A\) has twice the average power of \(B\)
ⓑ. Device \(B\) has twice the average power of \(A\), but their instantaneous powers at \(4\,\text{s}\) are equal
ⓒ. Their instantaneous powers are equal at all times
ⓓ. Their average powers are equal; at \(t=4\,\text{s}\), device B has twice the instantaneous power of A
Correct Answer: Their average powers are equal; at \(t=4\,\text{s}\), device B has twice the instantaneous power of A
Explanation: \(\textbf{Average power of A:}\)
At \(t=4\,\text{s}\),
\[
W_A=20(4)=80\,\text{J}
\]
Therefore
\[
P_{A,\text{avg}}=\frac{80}{4}=20\,\text{W}
\]
For device \(B\),
\[
W_B=5(4)^2=80\,\text{J}
\]
Hence
\[
P_{B,\text{avg}}=\frac{80}{4}=20\,\text{W}
\]
Their average powers over the complete interval are therefore equal.
\(\textbf{Instantaneous powers:}\)
For \(A\),
\[
P_A=\frac{dW_A}{dt}=20\,\text{W}
\]
For \(B\),
\[
P_B=\frac{dW_B}{dt}=10t.
\]
At \(t=4\,\text{s}\),
\[
P_B=40\,\text{W}=2P_A
\]
Equal total work over equal time does not require equal instantaneous power at each instant.
\(\textbf{Final answer:}\)
\[
\boxed{P_{A,\text{avg}}=P_{B,\text{avg}},\qquad P_B(4)=2P_A(4)}
\]
217. The cumulative work done by a machine is
\[
W(t)=at^2+bt,
\]
where \(W\) is in joules and \(t\) in seconds. Its average power over the first \(2\,\text{s}\) is \(10\,\text{W}\), while its instantaneous power at \(t=2\,\text{s}\) is \(14\,\text{W}\). Its instantaneous power at \(t=4\,\text{s}\) is
ⓐ. \(18\,\text{W}\)
ⓑ. \(20\,\text{W}\)
ⓒ. \(22\,\text{W}\)
ⓓ. \(28\,\text{W}\)
Correct Answer: \(22\,\text{W}\)
Explanation: The average-power and instantaneous-power conditions provide two independent equations for \(a\) and \(b\).
\(\textbf{Average power over }0\to2\,\text{s}\textbf{:}\)
\[
P_{\text{avg}}=\frac{W(2)-W(0)}{2}
\]
Thus
\[
10=\frac{4a+2b}{2}
\]
Hence
\[
2a+b=10
\]
\(\textbf{Instantaneous power at }2\,\text{s}\textbf{:}\)
\[
P(t)=\frac{dW}{dt}=2at+b
\]
Therefore
\[
14=4a+b
\]
Subtracting the first relation,
\[
2a=4,
\]
so
\[
a=2
\]
Then
\[
b=6
\]
At \(t=4\,\text{s}\),
\[
P(4)=2(2)(4)+6=22\,\text{W}
\]
\(\textbf{Final answer:}\)
\[
\boxed{22\,\text{W}}
\]
218. A machine has average power \(300\,\text{W}\) over its first \(4\,\text{s}\) of operation and average power \(420\,\text{W}\) over its first \(10\,\text{s}\). The work done from \(t=4\,\text{s}\) to \(t=10\,\text{s}\) and the average power during only this \(6\,\text{s}\) interval are
ⓐ. \(1800\,\text{J},\;300\,\text{W}\)
ⓑ. \(3000\,\text{J},\;500\,\text{W}\)
ⓒ. \(4200\,\text{J},\;420\,\text{W}\)
ⓓ. \(1200\,\text{J},\;200\,\text{W}\)
Correct Answer: \(3000\,\text{J},\;500\,\text{W}\)
Explanation: The two stated average powers refer to overlapping intervals, so the corresponding works must first be reconstructed.
\(\textbf{Work during the first }4\,\text{s}\textbf{:}\)
\[
W_{0\to4}=P_{\text{avg}}\Delta t
\]
\[
W_{0\to4}=(300)(4)=1200\,\text{J}
\]
\(\textbf{Work during the first }10\,\text{s}\textbf{:}\)
\[
W_{0\to10}=(420)(10)=4200\,\text{J}
\]
Therefore the work done between \(4\) and \(10\,\text{s}\) is
\[
W_{4\to10}=4200-1200=3000\,\text{J}
\]
The duration of this interval is
\[
10-4=6\,\text{s}
\]
Hence
\[
P_{4\to10,\text{avg}}=\frac{3000}{6}=500\,\text{W}
\]
The average over a subinterval cannot be found by simply subtracting the two quoted average powers.
\(\textbf{Final answer:}\)
\[
\boxed{3000\,\text{J},\;500\,\text{W}}
\]
219. Match List I with List II.
| List I | Quantity/representation | List II | Identification |
|---|
| P | \(\text{J s}^{-1}\) | 1 | Energy unit |
| Q | \(\text{kWh}\) | 2 | Power dimension |
| R | \([ML^2T^{-3}]\) | 3 | Watt |
| S | \(\text{hp}\) | 4 | Power unit convertible to \(746\,\text{W}\) |
Choose the correct matching.
ⓐ. P-3, Q-2, R-1, S-4
ⓑ. P-1, Q-3, R-4, S-2
ⓒ. P-3, Q-1, R-2, S-4
ⓓ. P-4, Q-1, R-3, S-2
Correct Answer: P-3, Q-1, R-2, S-4
Explanation: A watt is defined as
\[
1\,\text{W}=1\,\text{J s}^{-1},
\]
so P matches 3.
Kilowatt-hour is power multiplied by time and therefore represents energy, so Q matches 1.
The dimensions of power are
\[
[ML^2T^{-3}],
\]
so R matches 2.
Horsepower is another unit of power, with the conversion
\[
1\,\text{hp}=746\,\text{W}.
\]
Therefore S matches 4.
The required matching is
\[
\text{P-3,\ Q-1,\ R-2,\ S-4}.
\]
The key distinction is that kW and hp are power units, whereas kWh is an energy unit.
220. Four devices operate as shown below.
| Device | Power | Operating time |
|---|
| P | \(600\,\text{W}\) | \(20\,\text{min}\) |
| Q | \(1.2\,\text{kW}\) | \(10\,\text{min}\) |
| R | \(300\,\text{W}\) | \(30\,\text{min}\) |
| S | \(2\,\text{kW}\) | \(6\,\text{min}\) |
The correct decreasing order of energies transferred is
ⓐ. \(Q\gt S\gt P\gt R\)
ⓑ. \(P\gt Q\gt S\gt R\)
ⓒ. \(P=Q=S\gt R\)
ⓓ. \(S\gt Q\gt P=R\)
Correct Answer: \(P=Q=S\gt R\)
Explanation: It is convenient to calculate the energies in kilowatt-hour.
\(\textbf{Device P:}\)
\[
E_P=(0.6)\left(\frac{20}{60}\right)=0.20\,\text{kWh}.
\]
\(\textbf{Device Q:}\)
\[
E_Q=(1.2)\left(\frac{10}{60}\right)=0.20\,\text{kWh}.
\]
\(\textbf{Device R:}\)
\[
E_R=(0.3)\left(\frac{30}{60}\right)=0.15\,\text{kWh}.
\]
\(\textbf{Device S:}\)
\[
E_S=(2)\left(\frac6{60}\right)=0.20\,\text{kWh}.
\]
Therefore
\[
E_P=E_Q=E_S\gt E_R.
\]
A larger power rating does not necessarily imply a larger energy transfer because operating time also matters.
\(\textbf{Final answer:}\)
\[
P=Q=S\gt R.
\]
