Question:

A force acts on a 2 kg object so that its position is given as a function of time as x = 3t² + 5. What is the work done by this force in the first 5 seconds?

875 J

850 J

900 J

950 J

x = 3t² + 5
v = dx/dt
v = 6t + 0
At t = 0, v = 0
At t = 5 sec, v = 30 m/s
W.D. = ΔKE
W.D. = 1/2mv²
= 1/2(2)(30)²
= 900 J