First slide

Work done by Diff type of forces

Question

A force of F = 0.5 N is applied on lower block as shown in figure . The work done by lower block on upper block for a displacement of 3 m of the upper block with respect to ground is (Take, 9 = 10ms ²)

Difficult

A

$- 0.5 J$
B

$0.5 J$
C

2 J
D

$-$ 2J

Solution

Maximum acceleration of I kg block, $a_{\max} = μ g = 1 {ms}^{- 2}$
Common acceleration without relative motion between two blocks,, $a = \frac{0.5}{3} {ms}^{- 2}$
Since, $a < a_{\max}$
There will be no relative motion and blocks will move with acceleratlon $\frac{0.5}{3} {ms}^{- 2}$
Force of friction by lower block on upper block
$f = m a = (1) (\frac{0.5}{3}) = \frac{1}{6} N (towards right)$
Work done, $H = f \times s = 0.5 J$

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