# Frosty has a new pair of surfboards that are slippery. If Frosty’s mass is 50 kg and he moves on his surfboard at 25 m/s, what is needed to slow Frosty to a stop in 20 s?

Borys Shumyatskiy | Certified Educator

Hello!

I suppose Frosty slips without friction. In such a situation only two forces act on him, the gravity force downwards and the reaction force upwards. They are balanced and cannot stop him.

The only cause of velocity change of a body is a force (unbalanced). Suppose some constant force `F` will be applied to him in direction opposite to his movement. Then by the Newton's Second law he becomes to decelerate with the constant acceleration `a=F/m,` where `m` is the Frosty's mass.

For body moving straight with the initial velocity `V_0` and the constant (negative) acceleration `-a` its speed `V` may be found by the formula

`V(t)=V_0-a*t,`

where `t` is a time since the start. "Slow to a stop" means `V(t_1)=0,` so

`0=V_0-a*t_1,`

`V_0=a*t_1=(F/m)*t_1,`

therefore `F=V_0*m/t_1.`

`V_0,` `m` and `t_1` are given, so we can compute `F.`  It is `25*50/20=62.5(N).`

The answer: a constant force of the magnitude 62.5 N and the opposite direction is needed to stop Frosty in 20 s.