Question

A block of mass 5 kg is initially at rest
rough horizontal surface. A force of 45 acte
on it in a horizontal direction and pushes
over a distance of 2 m. The force of friction
acting on the block is 25 N. The final kinetic
energy of the block is​

Answer 1

Answer:

40 J

Explanation:

M=5kg, F=45N, s=2m, f=25N, kf=?

So, we have a formula fk=f x s

We have two forces friction and Force which are in opposite direction so we should subtract both the forces

F - f = 45 - 25

F=20N

fk=F x s

= 20 x 2

= 40 J

Hopes it helps u ☺️☺️