First slide

Work Energy Theorem

Question

A block of mass 2 kg slides down a curved track that is one quarter of a circle of radius 1 m [see fig. (1)]. Its speed at the bottom is 4 m,/s. Work done by frictional force is (take g = 10 m / s2 )

Moderate

A

-4J
B

8J
C

-8J
D

20J

Solution

In absence of friction, the velocity at the bottom of quarter circle
$V = \sqrt{2 gR} K.E. = \frac{1}{2} {mv}^{2} = mgR = 2 \times 10 \times 1 = 20 J Actual K.E. = \frac{1}{2} \times 2 \times (4)^{2} = 16 J Work done by frictional force = decrease in K E . = 16 J - 20 J = - 4 J$

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