Answer to Question #96900 in Mechanics | Relativity for Chelsea

Question

A block of mass 6.69 kg is on a horizontal surface. The coefficient of kinetic friction between the block and the surface is 0.159.
a) What horizontal force would be required to keep the block moving at a constant velocity?
b) What horizontal force is required to accelerate the block at 4.67 m/s^2 ?

Answer 1

The Newton's second law gives

$$ma=F-F_{\rm frict}$$ $$ma=F-\mu mg$$

(a) when a block miving with constant velocity its acceleration is zero , so

$$F=\mu mg=0.159\times 6.69\:\rm kg\times 9.81\:\rm m/s^2=10.4\: N$$

(b) if acceleration of the block is $4.67\: \rm m/s^2$, we obtain

$$F=ma+\mu mg$$ $$=6.69\:\rm kg\times4.67\: \rm m/s^2+10.4\: N =41.7\: N$$