A32kg 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                  

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