First slide

Numericals based on work energy theorem

Question

Under the action of a force, a 2 kg body moves such that its position x as a function of time t is given by x = $\frac{t^{3}}{3}$ , where x is in meter and t in seconds. The work done by the force in the first two seconds is:

Easy

A

1.6 J
B

16 J
C

160 J
D

1600 J

Solution

v = $\frac{dx}{dt} = \frac{d}{dt} (\frac{t^{3}}{3}) = t^{2}$
When t = 0, then v = o, When t = 2, then v = 4 m/s
Work done in first two seconds = change in KE
W = $\frac{1}{2} m [{(4)}^{2} - {(0)}^{2}] = \frac{1}{2} 2 \times 16 = 16$

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