A force acts on a 2 kg object so that its position is given as a function of time as x = 3t 2 + 5. What is the work done by this force in first 5 seconds?
Answers
Answered by
7
here is the answer of your math
Attachments:
Answered by
2
Thw the work done equals change in kinetic energy
W= 1\2mv^2 (change in kinetic energy)
M=2kg
Velocity equals differentiation of displacement wrt time=30m/s
Substitute the values and get the answer.
900joules.
9×10^9 ergs.
Similar questions