First slide

Concept of potential energy

Question

System shown in figure is released from rest. Pulley and spring are massless and friction is absent everywhere. The speed of 5kg block when 2kg block leaves contact with ground is $(K = 40 N m^{- 1}, g = 10 m s^{- 2})$ ______

Difficult

Solution

For 2kg: $2 g = k x \Rightarrow x = \frac{2 \times 10}{40} = \frac{1}{2} m$
Loss of P.E of 5kg block = gain of elastic potential energy of spring+ gain of kinetic energy of 5kg
$\begin{array}{l} 5 \times g x = \frac{1}{2} k x^{2} + \frac{1}{2} m v^{2} \\ 5 \times 10 \times \frac{1}{2} = \frac{1}{2} \times 40 \times \frac{1}{4} + \frac{1}{2} \times 5 \times v^{2} \\ 4 = \frac{v^{2}}{2} \\ v = 2 \sqrt{2} = 2 \times 1.414 = 2.828 {ms}^{- 1} \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App