Calculate the work done by this force on the mass

Question

1) A force, F = 43.5Nî + 28.0Nj is exerted on a mass, m = 12.0kg and moves in the x-direction a distance of 15.0m.

2) Determine the angle between the force and displacement vectors. Use the dot product to find the angle.

Answer 1

By definition, the work done by the force F along a path γ is equal to the integral along γ of F·dr (dot product between the force and the displacement) where here in bold are indicate vector quantities. In our case we have F=(F_x,F_y) and dr=(dx,dy).

Given this, the answer to the first question is given by evaluating the integral between 0 (arbitrary initial point) and 15 (final point) of F_x =43.5, which is given simply by the product between the x-component of the force and the displacement. Hence, the result is W=43.5N*15m=652.5 J

The angle between two vectors is given by the ratio between the dot product of the two divided by the product of the magnitude of the two: cos(α)=F·dr/|F|·|dr| where the vertical bars indicate the magnitude.

In this case we have F=(F_x,F_y)=(43.5,28) and dr=(15,0) and hence

F·dr=(43.5,28)·(15,0)=43.5*15=652.5

|F|=sqrt(43.5²+ 28²)≈51.73

|dr|=sqrt(15²+ 0²)=15

cos(α)=625.5/(51.73*15)≈0.84

and thus, by inverting the above you have that α=arccos(0.84)≈32.85 degrees.

Best,

Davide