A single nonconstant force acts in the x direction on an 506

A single, non-constant force acts in the +x direction on an 5.06 kg object that is constrained to move along the x axis. As a result the object\'s position as a function of time is: x(t) = A + Bt + Ct^3 A = 6.43 m B = 4.37 m/s C = 0.158 m/s^3 How much work is done by this force from t = 0 s to t = 2.64 s?

Solution

We need to determine the work done by the given force between the time interval of t = 0 to t = 2.64 seconds.

Now, we know that the work done on a body is given as: W = F(x).dx

That would mean that we will have to determine the expression for the force in terms of time t and then get a product of force and displacement. We will then need to integrate this product over the given time interval to get the required value.

We have: x(t) = A + Bt + Ct³

Now, velocity = dx/dt = B + 3Ct²

also, acceleration = dv/dt = 6Ct

Therefore the force = M x acceleration = 30.36Ct

So we can write the expression for work as dW = F.dx

or, dW = 30.36Ct (B + 3Ct²) dt

We integrate the above expression for t from 0 to 2.64 to get total work done as:

W = dW = 30.36Ct (B + 3Ct²) dt

or, W = 30.36C [Bt²/2 + 3Ct⁴/4]

or, W = 30.36 x 0.158 [4.37(2.64)²/2 + 3(0.158)(2.64)⁴/4]

or, W = 30.36 x 0.158 [5.75618 + 15.2286] = 100.6614 Joules is the required work done.