A block of ice with mass 6.10kg is initially at rest on a frictionless, horizont

Question

A block of ice with mass 6.10kg is initially at rest on a frictionless, horizontal surface. A worker then applies a horizontal force F to it. As a result, the block moves along the x-axis such that its position as a function of time is given by x(t)=(0.192m/s^2)t^2 + ( 2.05*10^?2 m/s^3)t^3.

a) Calculate the velocity of the object at time t=3.80s
I solved this one and the answer is 2.35 m/s

b) Calculate the magnitude of F at time t = 3.80s.
I solved this one and the answer is 5.19N

c) Calculate the work done by the force F during the first time interval of 3.80s of the motion.

so far I've gotten to

F*v = ((2)(0.192 m/s^2) + (6)(2.05

Explanation / Answer

c) work done by the force F = force*displacement = F*x(t) F u already done so find x(t) at t=3.8s then u vl get d answer

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