Work Energy And Power Question 64
Question: The displacement x of a particle moving in one dimension under the action of a constant force is related to the time t by the equation $ t=\sqrt{x}+3 $ , where x is in meters and t is in seconds. The work done by the force in the first 6 seconds is [IIT 1979]
Options:
A) 9 J
B) 6 J
C) 0 J
D) 3 J
Show Answer
Answer:
Correct Answer: C
Solution:
$ x={{(t-3)}^{2}} $
therefore $ v=\frac{dx}{dt}=2(t-3) $
at $ t=0 $ ; $ v_1=-6m/s $ and at $ t=6\sec $ , $ v_2=6m/s $
so, change in kinetic energy $ =W=\frac{1}{2}mv_2^{2}-\frac{1}{2}mv_1^{2}=0 $
BETA
