Given that,
Mass of a box, m = 150 kg
It is pushed at an angle of 15 degrees with the horizontal.
It is 8.1 m long.
The coefficient of kinetic friction between the box and the incline is 0.4
To find,
The work done in pushing the box.
Solution,
First let's see the net force acting on the box.
[tex]F_{net}=mg\sin\theta+f_k\\=mg\sin\theta+\mu_k mg\cos\theta[/tex]
Work done is given by :
W = force × displacement
[tex]W=(mg\sin\theta+\mu_k mg\cos\theta)\times s\\\\=150\times 10(\sin15+8.1\times \cos15)\times 8.1\\\\=98206.24\\\\or\\\\=98.2\ kJ[/tex]
So, the required work done is 98.2 kJ.
