Question

A single, nonconstant force acts in the +x-+x-direction on a 4.06 kg4.06 kg object that is constrained to move along the x-x-axis. As a result, the object's position as a function of time is

x(t)=A+Bt+Ct^3

where A=4.43 m, B=3.12 m/s, and C=0.308 m/s.

How much work is done by this force from t=0 st=0 s to t=2.14 s?

Answer 1

The force acting on the object
F= m a = m dv/dt = m d2x/dt2
F = m d2(A + B t + C t3) /dt2
F = m d (B + 3 C t2)/dt
F = m . 6 C t
F = 4.06 kg x 6 x 0.308 m/s x t
F = 7.503 t
F is the function of time
F(t) = 7.503 t
The work done is
W = F. x
Where x is the displacement

Where v is the velocity of the object,
v = d x / dt = B + 3 C t2, Subtituing this,


Substituting the values for A, B and C we get
W = 553.57 J