First slide

Work done by Diff type of forces

Question

In the arrangement shown, on external agent slowly pulls the block P from A to B along the rough inclined surface AB. The coefficient of friction between the block and the wedge is 0.75. Then the ratio of changes in potential energy of the block to the work done by the agent is

Difficult

A

4 : 5
B

2 : 7
C

4 : 7
D

3 : 5

Solution

(3)
$\begin{array}{l} Δ U = C h a n g e i n p o t e n t i a l e n e r g y = m g h \\ w_{f} = w o r k d o n e a g a i n s t f r a c t i o n a l f o r c e \\ = f \times A B = μ m g \cos θ \times \frac{h}{\sin θ} = μ m g h . \cot θ \\ ∴ w = w o r k d o n e b y e x t e r n a l a g e n t = w_{f} + Δ U = m g h (1 + μ \cos θ) \\ ∴ \frac{Δ U}{w} = \frac{m g h}{m g h (1 + μ \cot θ)} = \frac{1}{(1 + \frac{3}{4} \cot 45^{0})} = \frac{4}{7} \end{array}$

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