Numerical Questions-2 - Chapter 4 Work Energy and Power 11th Science Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Numerical Questions
Question 1.
A body is moving along $z$-axis of a coordinate system under the effect of a constant force $F=F$ ind the work done by the force in moring the body a distance of $2 \mathrm{~m}$ along $z$-axis.
Answer:
$
\begin{aligned}
\overrightarrow{\mathrm{F}} & =(2 \hat{i}+3 \hat{j}+\hat{k}) \mathrm{N}, \overrightarrow{\mathrm{S}}=2 \hat{k} \\
\mathrm{~W} & =\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{S}}=2 \mathrm{~J}
\end{aligned}
$

Question 2.
Water is pumped out of a well $10 \mathrm{~m}$ deep by means of a pump rated $10 \mathrm{KW}$. Find the efficiency of the motor if $4200 \mathrm{~kg}$ of water is pumped out every minute. Take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s} 2$.
Answer:
Input power $=10 \mathrm{KW}$
Output power $=\frac{\mathrm{W}}{t}=\frac{m g h}{t}=7 \mathrm{KW}$
$\therefore$ Efficiency $=\frac{\text { Output power }}{\text { Input power }} \times 100=70 \%$
Question 3.
A railway carriage of mass $9000 \mathrm{~kg}$ moving with a speed of $36 \mathrm{kmph}$ collides with a stationary carriage of same mass. After the collision, the carriages get coupled and move together. What is their common speed after collision? What type of collision is this?
Answer:
$
\begin{aligned}
& \mathrm{m}_1=9000 \mathrm{~kg}, \mathrm{u}_1=36 \mathrm{~km} / \mathrm{h}=10 \mathrm{~m} / \mathrm{s} \\
& \mathrm{m}_2=9000 \mathrm{~kg}, \mathrm{u}_2=0, \mathrm{v}=\mathrm{v}_1=\mathrm{v}_2=?
\end{aligned}
$
By conservation of momentum:
$\mathrm{m}_1 \mathrm{u}_1+\mathrm{m}_2 \mathrm{u}_2=\left(\mathrm{m}_1+\mathrm{m}_2\right) \mathrm{v}$
$
\therefore \mathrm{v}=5 \mathrm{~m} / \mathrm{s}
$
Total K.E. before collision $=\frac{1}{2} m_1 u_1^2+\frac{1}{2} m_2 u_2^2=45000 \mathrm{~J}$
'Total K.E. after collision $=\frac{1}{2}\left(m_1+m_2\right) v^2=225000 \mathrm{~J}$
As total K.E. after collision $<$ Total K.E. before collision
$\therefore$ collision is inelastic
Question 4.
In lifting a $10 \mathrm{~kg}$ weight to a height of $2 \mathrm{~m}, 230 \mathrm{~J}$ energy is spent. Calculate the acceleration with which it was raised.

Answer:
$
\begin{aligned}
& \mathrm{W}=\mathrm{mgh}+\mathrm{mah}=\mathrm{m}(\mathrm{g}+\mathrm{a}) \mathrm{h} \\
& \therefore \mathrm{a}=1.5 \mathrm{~m} / \mathrm{s}^2 .
\end{aligned}
$
Question 5.
A bullet of mass $0.02 \mathrm{~kg}$ is moving with a speed of $10 \mathrm{~ms}^{-1}$. It can penetrate $10 \mathrm{~cm}$ of a wooden block, and comes to rest. If the thickness of the target would be $6 \mathrm{~cm}$ only, find the K.E. of the bullet when it comes out.
For $x=10 \mathrm{~cm}=0 \cdot 1 \mathrm{~m}, \mathrm{~F} x=\frac{1}{2} m v_1^2=1 \mathrm{~J}$
$
\therefore \quad \mathrm{F}=10 \mathrm{~N}
$
For $x=6 \mathrm{~cm}=0.06 \mathrm{~m}, \mathrm{~F} x=\frac{1}{2} m v_1^2-\frac{1}{2} m v_2^2$.
or
$
\mathrm{F} x=\frac{1}{2} m v_1^2-\text { Final K.E. }
$
or
$
\text { Final K.E. }=\frac{1}{2} m v_1^2-\mathrm{F} x=1-10 \times 0.06=1-0.6=0.4 \mathrm{~J}
$
Question 6.
A man pulls a lawn roller through a distance of $20 \mathrm{~m}$ with a force of $20 \mathrm{~kg}$ weight. If he applies the force at an angle of $60^{\circ}$ with the ground, calculate the power developed if he takes $1 \mathrm{~min}$ in doing so.

Answer:
$
\mathrm{P}=\frac{\mathrm{W}}{t}=\frac{\mathrm{Fs} \cos \theta}{t}=32.66 \mathrm{~W} \text {. }
$
Question 7.
A body of mass $0.3 \mathrm{~kg}$ is taken up an inclined plane to length $10 \mathrm{~m}$ and height $5 \mathrm{~m}$ and then allowed to slide down to the bottom again. The coefficient of friction between the body and the plane is 0.15 . What is the
(i) work done by the gravitational force over the round trip?
(ii) work done by the applied force over the upward journey?
(iii) work done by frictional force over the round trip?
(iv) kinetic energy of the body at the end of the trip?
How is the answer to (iv) related to the first three answers?
Answer:
$
\begin{aligned}
\sin \theta & =\frac{\mathrm{CB}}{\mathrm{CA}}=0.5 \\
\theta & =30^{\circ}
\end{aligned}
$

(i) $\mathrm{W}=\mathrm{FS}=-\mathrm{mg} \sin \theta \times \mathrm{h}=-14.7 \mathrm{~J}$ is the W.D. by gravitational force in moving plane. $\mathrm{W}^{\prime}=\mathrm{FS}=+\mathrm{mg} \sin \theta \times \mathrm{h}=14.7 \mathrm{~J}$ is the W.D. by gravitational force in moving the body down the inclined plane.
$\therefore$ Total W.D. round the trip, $\mathrm{W}_1=\mathrm{W}+\mathrm{W}^{\prime}=0$
(ii) Force needed to move the body up the inclined plane, $\mathrm{F}=\mathrm{mg} \sin \theta+\mathrm{f}_{\mathrm{k}}=\mathrm{mg} \sin \theta+\mu_{\mathrm{k}} \mathrm{R}=\mathrm{mg} \sin \theta+\mu_{\mathrm{k}} \mathrm{mg} \cos \theta$
$\therefore$ W.D. by force over the upward journey is $\mathrm{W}_2=\mathrm{F} \times 1=\mathrm{mg}\left(\sin \theta+\mu_{\mathrm{k}} \cos \theta\right) 1=18.5 \mathrm{~J}$
(iii) W.D. by frictional force over the round trip, $\mathrm{W}_3=-\mathrm{fk}(1+1)=-2 \mathrm{fkl}=-2 \mu_{\mathrm{k}} \cos \theta 1=-7.6 \mathrm{~J}$
(iv) K.E. of the body at the end of round trip
$=$ W.D. by net force in moving the body down the inclined plane
$=\left(\mathrm{mg} \sin \theta-\mu_{\mathrm{k}} \cos \theta\right) 1$
$=10.9 \mathrm{~J}$
$\Rightarrow$ K.E. of body $=$ net W.D. on the body.
Question 8.
Two identical $5 \mathrm{~kg}$ blocks are moving with same speed of $2 \mathrm{~ms}^{-1}$ towards each other along a frictionless horizontal surface. The two blocks collide, stick together and come to rest. Consider the two blocks as a system. Calculate work done by
(i) external forces and
(ii) Internal forces.
Answer:
Here no external forces are acting on the system so:
$\overrightarrow{\mathrm{F}}_{\text {ext }}=0 \Rightarrow \mathrm{W}_{\text {ext }}=0$
According to work-energy theorem :
Total W.D. = Change in K.E.
or $\mathrm{W}_{\text {ext }}+=$ Final K.E. - Initial K.E.
or
$
\begin{aligned}
0+\mathrm{W}_{\text {int. }} & =0-\left(\frac{1}{2} m u^2+\frac{1}{2} m u^2\right) \\
\mathrm{W}_{\text {int. }} & =-m u^2=-20 \mathrm{~J}
\end{aligned}
$

Question 9.
A truck of mass $1000 \mathrm{~kg}$ accelerates uniformly from rest to a velocity of $15 \mathrm{~ms}^{-1}$ in 5 seconds. Calculate (i) its acceleration,
(ii) its gain in K.E.,
(iii) average power of the engine during this period, neglect friction.
Answer:
(i) $a=\frac{v-u}{t}=3 \mathrm{~m} / \mathrm{s}^2$
(ii) Gain in K.E. $=\frac{1}{2} m\left(v^2-u^2\right)=1.125 \times 10^5 \mathrm{~J}$
(iii) $\mathrm{P}=\frac{\mathrm{W}}{t}=22500 \mathrm{~W}$.
Question 10.
An elevator which can carry a maximum load of $1800 \mathrm{~kg}$ (elevator + passengers) is moving up with a constant speed of $2 \mathrm{~ms}^{-1}$. The frictional force opposing the motion is $4000 \mathrm{~N}$. Determine the minimum
power delivered by the motor to the elevator in watts as well as in horse power.
Answer:
Downward force on the elevator is :
$\mathrm{F}=\mathrm{mg}+\mathrm{f}=22000 \mathrm{~N}$
$\therefore$ Power supplied by motor to balance this force is :
$
\mathrm{P}=\mathrm{F} v=44000 \mathrm{~W}=\frac{44000}{746}=59 \mathrm{hp} .
$
Question 11.
To simulate car accidents, auto manufacturers study the collisions of moving cars with mounted springs of different spring constants. Consider a typical simulation with a car of mass $1000 \mathrm{~kg}$ moving with a speed $18.0 \mathrm{kmh}^{-1}$ on a smooth road and colliding with a horizontally mounted spring of spring constant $6.25 \times 10^{-3} \mathrm{Nm}^{-1}$. What is the maximum compression of the spring?
Answer:
At maximum compression $\mathrm{x}_{\mathrm{m}}$, the K.E. of the car is converted entirely into the P.E. of the spring.

$\therefore \quad \frac{1}{2} k x_m^2=\frac{1}{2} m v^2 \text { or } x_m=2 \mathrm{~m} .$