
A mall mass m is suspended on a weightless elastic thread whose stiffness factor is k. The ball is lifted so that the thread is not stretched and let fall with zero initial velocity. The maximum stretch of the thread in the process of ball’s motion is
(A) $ \dfrac{\text{mg}}{\text{k}} $
(B) $ \dfrac{2\text{ mg}}{\text{k}} $
(C) $ -\dfrac{\text{G}{{\text{M}}^{1}}}{\text{H}} $
(D) $ -\dfrac{6}{5}\dfrac{\text{G}{{\text{M}}^{2}}}{\text{H}} $
Answer
446.7k+ views
Hint: Stiffness factor is the ratio of moment of inertia cross-section of beam or wire to its length. It is a measure of “resistance to bending”. Kinetic energy of a body is the energy possessed by the body by virtue of its motion. It is given by:
$ \text{E}=\dfrac{1}{2}\text{m}{{\text{v}}^{2}} $
Gravitational potential energy of a body is the energy possessed by the body by virtue of its position above the surface of the earth.
E = mgh
Complete step by step solution
Consider a ball of mass ‘m’ is suspended on a weightless elastic thread whose stiffness factor is k. The ball is lifted so that the thread is not stretched and let fall with zero initial velocity.
We will find out the maximum stretch of the thread in the process of ball’s motion i.e. ‘x’ we know that potential energy is stored in the thread.
It is given by: $ \text{E}=\dfrac{1}{2}\text{k }{{\text{x}}^{2}} $ ….. (1)
And, we also know that decrease in gravitational P.E
E = mgx …. (2)
To find out the maximum stretching the equation (1) should be equal to (2)
$ \therefore \text{ mgx}=\dfrac{1}{2}\text{k}{{\text{x}}^{2}} $
$ \text{mg}=\dfrac{1}{2}\text{kx} $
$ \text{x}=\dfrac{2\text{ mg}}{\text{k}} $
So, the maximum stretch of the thread in the process of ball’s motion is
$ \text{x}=\dfrac{2\text{ mg}}{\text{k}} $
Therefore
Option (B) is correct.
Note
Do not confuse yourself with spring and elastic thread. A spring will resist compression but a string will not compression. Keep in mind that tension force and spring force are different things. Tension force is the force that is transmitted through a string, rope but the spring force is the force exerted by a compressed or stretched spring upon any object that is attached to it.
$ \text{E}=\dfrac{1}{2}\text{m}{{\text{v}}^{2}} $
Gravitational potential energy of a body is the energy possessed by the body by virtue of its position above the surface of the earth.
E = mgh
Complete step by step solution
Consider a ball of mass ‘m’ is suspended on a weightless elastic thread whose stiffness factor is k. The ball is lifted so that the thread is not stretched and let fall with zero initial velocity.
We will find out the maximum stretch of the thread in the process of ball’s motion i.e. ‘x’ we know that potential energy is stored in the thread.
It is given by: $ \text{E}=\dfrac{1}{2}\text{k }{{\text{x}}^{2}} $ ….. (1)
And, we also know that decrease in gravitational P.E
E = mgx …. (2)
To find out the maximum stretching the equation (1) should be equal to (2)
$ \therefore \text{ mgx}=\dfrac{1}{2}\text{k}{{\text{x}}^{2}} $
$ \text{mg}=\dfrac{1}{2}\text{kx} $
$ \text{x}=\dfrac{2\text{ mg}}{\text{k}} $
So, the maximum stretch of the thread in the process of ball’s motion is
$ \text{x}=\dfrac{2\text{ mg}}{\text{k}} $
Therefore
Option (B) is correct.
Note
Do not confuse yourself with spring and elastic thread. A spring will resist compression but a string will not compression. Keep in mind that tension force and spring force are different things. Tension force is the force that is transmitted through a string, rope but the spring force is the force exerted by a compressed or stretched spring upon any object that is attached to it.
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