A 32kg kid on a sled sliding over some rough snow is acted on by a frictional force which reduces its velocity form 8.5 m/s to 4.1 m/s in 3.0 seconds what is the magnitude of the frictional force?

F = 47.04 N

Explanation:

Given,

The mass of the kid, m = 32 Kg

The initial velocity of the kid, u = 8.5 m/s

The final velocity of the kind, v = 4.1 m/s

The time period of travel. t = 3 s

Using the first equations of motion,

v = u + at

a = (v - u) / t

= (4.1 - 8.5) / 3

= - 1.47 m/s²

The negative sign indicates that the acceleration of the frictional force is against the motion of the child.

The acceleration of the frictional force, a = 1.47 m/s²

Therefore, the frictional force,

F = m x a

= 32 x 1.47

= 47.04 N

Hence, the frictional force acting on the child is, F = 47.04 N