Question

When a force of 500 N pushes on a 25 kg box as shown in figure, the acceleration of the box up the incline is $0.75m/s^2$. Find the coefficient of kinetic friction between box and inclined plane .

Answer 1

Given: when a force of 500 N pushes on a 25 kg box as shown in figure, the acceleration of the box up the incline is $0.75m/s^2$.

To find the coefficient of kinetic friction between box and inclined plane

Solution:

From figure,

$ma+mg\sin h_0+f=F\therefore \left(25\right)\left(0.75\right)+\left(25\right)\left(10\right)\sin h_0+{\mu }_{k}N=F\therefore 18.75+\left(250\right)\left(0.6h\right)+{\mu }_k(mg\cos h_0=F⟹18.75+150+{\mu }_k(\left(25\right)\left(10\right)\left(0.76\right))=500\therefore 168.75+{\mu }_k(190)=500⟹{\mu }_k(190)=331.25⟹{\mu }_k=1.74$

is the coefficient of kinetic friction between box and inclined plane