3 Marks Question

Question 13 Marks

In an ideal spring the law $\vec{F}=-k \vec{x}$ is followed. Here $k$ is the spring constant. Calculate the work done by the spring in stretching it to a distance $x$ from its equilibrium position by integral method and graphical method

Answer

Integration method :
$\begin{array}{l}
\text { Work } \quad w=\int_0^x \overrightarrow{F} \cdot \overrightarrow{d x} \\
\because \quad \vec{F}=-k \vec{x} \\
\therefore \quad w=\int_0^x-k \vec{x} \cdot \overrightarrow{d x} \\
w=-k \int_0^x x d x=-k\left[\frac{x^2}{2}\right]_0^x \\
=-k \frac{x^2}{2}=-\frac{1}{2} k x^2
\end{array}$
Graphical method : If a graph is drawn between the increase in spring length $x$ and the force F , then a straight line is obtained as shown in figure. Work done by restoring force :

$\begin{aligned}
w & =\text { area of } \Delta OPQ \\
w & =\frac{1}{2} \times OP \times PQ \\
& =\frac{1}{2} x \times(-k x) \\
& =-\frac{1}{2} k x^2
\end{aligned}$
Hence, the value of work obtained by both methods is same.

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Question 23 Marks

Prove that as the object moves towards the ground from some height, its potential energy decreases and kinetic energy increases by the same amount.

Answer

Suppose an object of mass $m$ is at rest at point A at a height $h$ from the earth's surface as shown in figure. At point A, object will have only potential energy and its kinetic energy will be zero. Value of potential energy at point $A = mgh$. Suppose the object falls freely and when the object reaches the ground, its speed is $v$ hence
the value of kinetic energy at point B will be $\frac{1}{2} m v^2$. Here, $h$ $=0$ hence potential energy $U =0$.

From the third equation of motion the speed of object at point $B$ is
$\begin{aligned}
v^2 & =u^2+2 a s \\
v^2 & =0+2 g h \\
\therefore \quad v^2 & =2 g h
\end{aligned}$
$\therefore \quad$ Kinetic energy at point B
$\begin{array}{l}
K=\frac{1}{2} m v^2 \\
K=\frac{1}{2} m \times 2 g h=m g h \\
K=m g h
\end{array}$
That is, the value of kinetic energy of an object on earth's surface is equal to the potential energy at point A . The decrease in potential energy of a freely following object is equal to increase in its kinetic energy. As the objects comes from height towards the ground the potential energy decreases and the kinetic energy increases by the same amount.
$\Delta K=-\Delta U$

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Question 33 Marks

Write the main characteristics of potential energy

Answer

(i) Potential energy is a scalar function of position.
(ii) The negative value of the derivative of potential energy with respect to position determines the force $\overrightarrow{F}=-\frac{d}{d t} U_{(x)}$
(iii) The value of potential energy depends on the frame of reference.
(iv) The concept of potential energy is defined only for internal conservative forces and not for non-conservative forces.
(v) Change in potential energy continous during the motion of a rigid object because the distance between its particles doesn't change even when external force is applied to the motion of a rigid object.

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Question 43 Marks

What is work and also give examples of the following :(i) Zero work(ii) Positive work(iii) Negative work

Answer

Suppose a constant force F is acting on an object of mass $m$ due to which the displacement of the
object in the positive $x$-direction is $\overrightarrow{ S }$, as shown in the figure.

Therefore the work done by a force is defined as 'the product of the magnitude of displacement and component of force along the direction of displacment. Therefore,
$
W=(F \cos \theta) s=\overrightarrow{F} \cdot \overrightarrow{S} \ \ldots(1)
$
(i) Zero work : It is clear from equation (1) that if the displacement of the object is zero then no matter how great is the magnitude of force, the work done by the object is zero. No work is considered completed if
(a) the displacement is zero
(b) force is zero
(c) force and displacement are perpendicular to each other. Since $\theta=\frac{\pi}{2}$ and $\cos \frac{\pi}{2}=0$ is always zero.
The centripetal force on an object moving in a circular path with constant speed is perpendicular to the displacement hence the work done by the centripetal force is zero.
(ii) Positive work : If $\theta \geq 0^{\circ}$ and $\theta<90^{\circ}$ then the value of $\cos \theta$ is positive. In such a situation the work is positive. For example, when a man pulls a roller, the value of work is positive.
(iii) Negative work: When $90^{\circ}<\theta \leq 180^{\circ}$ then the value of $\cos \theta$ is negative. Hence the work is negative.
For example, when a roller is pulled on a rough surface, the friction force and displacement are in opposite directions hence the work done by the friction force is negative.

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