Given :
A weight lifter lifts a 280-N set of weights from ground level to a position over his head, a vertical distance of 1.60 m.
To Find :
How much work does the weight lifter do, assuming he moves the weights at constant speed.
Solution :
We know, work done is given by :
[tex]W= Fdcos\ \theta[/tex]
Here, F = 280 N and d = 1.60 m
Since, weight lifer is applying and upward force and displacement is also in upward direction.
[tex]\theta = 0^{\circ}\\cos\ 0^{\circ} = 1[/tex]
Putting all these in above equation, we get :
[tex]W = 280\times 1.6\times 1\\\\W = 448\ J[/tex]
Therefore, work done by weight lifter 448 J.
Hence, this is the required solution.
