First slide

Work done by Diff type of forces

Question

A force of (4x² + 3x) N acts on a particle which displaces it from x = 2m to x = 3m. The work done by the force is

Moderate

A

32.8J
B

3.28J
C

0.328J
D

zero

Solution

$\large W = \int\limits_2^3 {Fdx} = 4\left[ {\frac{{{x^3}}}{3}} \right]_2^3 + 3\left[ {\frac{{{x^2}}}{2}} \right]_2^3$
$\large W = \frac{4}{3}(27 - 8) + \frac{3}{2}(9 - 4)\;;\;W = 32.8J$

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