Question

The pulley shown in figure (10-E11) has a radius of 20 cm and moment of inertia 0⋅2 kg-m2. The string going over it is attached at one end to a vertical spring of spring constant 50 N/m fixed from below, and supports a 1 kg mass at the other end. The system is released from rest with the spring at its natural length. Find the speed of the block when it has descended through 10 cm. Take g = 10 m/s2.

Answer 1

The speed of the block is 0.5 m/s.

Explanation:

Given that,

Radius = 20 cm

Moment of inertia = 0.2 kg m²

Spring constant = 50 N/m

Mass = 1 kg

Height = 10 cm = 0.1 m

We need to calculate the velocity of the block

Using law of conservation of energy

$mgh=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2+\dfrac{1}{2}kx^2$

$mgh=\dfrac{1}{2}mv^2+\dfrac{1}{2}I(\dfrac{v}{r})^2+\dfrac{1}{2}kx^2$

Put the value into the formula

$1\times10\times10\times10^{-2}=\dfrac{1}{2}\times 1 v^2+\dfrac{1}{2}\times0.2\times(\dfrac{v^2}{(20\times10^{-2})^2})+\dfrac{1}{2}\times50\times(10\times10^{-2})^2$

$1=0.5v^2+2.5v^2+0.25$

$3.0v^2=0.75$

$v^2=\dfrac{0.75}{3.0}$

$v=\sqrt{\dfrac{0.75}{3.0}}$

$v=0.5\ m/s$

Hence, The speed of the block is 0.5 m/s.

Learn more :

Topic : energy conservation

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