A 48.8-kg skater is standing at rest in front of a wall. By pushing against the wall she propels herself backward with a velocity of -1.88 m/s. Her hands are in contact with the wall for 0.763 s. Ignore friction and wind resistance. Find 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). Note that this force has direction, which you should indicate with the sign of your answer.
To find the average force exerted by the skater on the wall, we can use Newton's second law of motion, which states that force is equal to mass multiplied by acceleration (F = m * a).
In this case, the skater starts at rest (so her initial velocity, u, is 0) and ends up with a final velocity, v, of -1.88 m/s. The time taken (t) is given as 0.763 s.
We can find the acceleration of the skater by using the equation:
a = (v - u) / t
Substituting the values, we get:
a = (-1.88 m/s - 0) / 0.763 s
= -2.463 m/s^2
The negative sign indicates that the acceleration is in the opposite direction of the velocity. Now, we can calculate the force applied by the skater:
F = m * a
= 48.8 kg * -2.463 m/s^2
= -119.9184 N
Therefore, the average force exerted by the skater on the wall is approximately -119.92 N. The negative sign indicates that the force is in the opposite direction of the skater's motion, as mentioned in the question.