First slide

Work

Question

A smaller block of 2kg is placed over a larger block of 5kg. Coefficient of friction between blocks is 0.3 and large block is on smooth surface. A force of 2N is applied on lower block, then work done by friction on upper block in 3 seconds is

Very difficult

Solution

$\begin{array}{l} f_{l} = μ m_{1} g = 0.3 \times 2 \times 10 = 6 N \\ F < f_{l} \\ \Rightarrow blocks move together \\ a = \frac{F}{m_{1} + m_{2}} = \frac{2}{7} {ms}^{- 2} \\ f = m a = f = 2 \times \frac{2}{7} = \frac{4}{7} N \\ W = {\vec{f}}_{1} \cdot \vec{S_{1}} = \frac{4}{7} \times \frac{1}{2} \times \frac{2}{7} \times 3^{2} = \frac{36}{49} = 0.735 J \end{array}$

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