Question

A coin slides over a frictionless plane and across an xy coordinate system from the origin to a point with xy coordinates (4.0 m, 5.8 m) while a constant force acts on it. The force has magnitude 1.3 N and is directed at a counterclockwise angle of 121° from the positive direction of the x axis. How much work is done by the force on the coin during the displacement?

I understand that W=Fd and F=MA

W=1.3*d, but Im not quite sure how exactly to find displacement.

Answer 1

Answer:

Work done, W = 3.78 Joules

Explanation:

It is given that,

Displacement in x-y coordinate, $d=(4i+5.8j)\ m$

Magnitude of force, F = 1.3 N

The force is directed at a counterclockwise angle of 121° from the positive direction of the x axis. Let W is the work done by the force on the coin during the displacement.

Firstly, finding the vector form of force as :

$F=1.3\ cos(121)i+1.3\ sin(121)j$

$F=-0.669i+1.114j$

The work done is given by :

$W=F.d$

$W=(-0.669i+1.114j).(4i+5.8j)$

$W=-0.669\times 4+1.114\times 5.8$

W = 3.78 Joules

So, the work done by the force is 3.78 Joules. Hence, this is the required solution.