Question

A 58-kg skater is standing still in front of a wall. By pushing against the wall she propels herself backward with a velocity of -1 m/s. Her hands are in contact with the wall for 0.80 s. Ignore friction and wind resistance. Find the magnitude and direction of the average force she exerts on the wall (which has the same magnitude, but opposite direction, as the force that the wall applies to her).

Answer 1

Answer:

F = -72.5 N

Explanation:

You can do this excercise in two ways.

The first way is using the second law of Newton which is:

F = m*a

You should calculate the acceleration and then solve for F.

The second way is to use the formula of momentum which is:

F*t = m*V

Here, you only need to put the data of the excercise and solve for F.

Either way you do this, you'll get the same result.

I'll do it by both methods to prove it:

First method:

Assuming that the innitial speed is 0 (because is standing still) acceleration would be:

a = Vf/t

Solving for a:

a = -1/0.8 = -1.25 m/s^2

Now, let's solve for F:

F = 58 * (-1.25) = -72.5 N

Second method:

We know that F*t = m*V so:

F = m*V/t

F = 58 * (-1) / 0.8

F = -72.5 N