Notes: 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...

Tuesday, 15 September 2015

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...

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 constantforce $\displaystyle{F}$ 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...

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 $\displaystyle{F}$ 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 $\displaystyle{a}=\frac{{F}}{{m}},$ where $\displaystyle{m}$ is the Frosty's mass.

For body moving straight with the initial velocity $\displaystyle{V}_{{0}}$ and the constant (negative) acceleration $\displaystyle-{a}$ its speed $\displaystyle{V}$ may be found by the formula

$\displaystyle{V}{\left({t}\right)}={V}_{{0}}-{a}\cdot{t},$

where $\displaystyle{t}$ is a time since the start. "Slow to a stop" means $\displaystyle{V}{\left({t}_{{1}}\right)}={0},$ so

$\displaystyle{0}={V}_{{0}}-{a}\cdot{t}_{{1}},$

$\displaystyle{V}_{{0}}={a}\cdot{t}_{{1}}={\left(\frac{{F}}{{m}}\right)}\cdot{t}_{{1}},$

therefore $\displaystyle{F}={V}_{{0}}\cdot\frac{{m}}{{t}_{{1}}}.$

$\displaystyle{V}_{{0}},$ $\displaystyle{m}$ and $\displaystyle{t}_{{1}}$ are given, so we can compute $\displaystyle{F}.$ It is $\displaystyle{25}\cdot\frac{{50}}{{20}}={62.5}{\left({N}\right)}.$