21) A 46 kg skater is standing still in front of a wall. By pushing against the wall she propels herself backward with a velocity of – 1.2 m /s. Her hands are in contact with the wall for 0.8 s. Ignore friction and wind resistance
• Find the magnitude and direction of the average force she exerts on the wall (which is the same magnitude, but in the other direction, as the force that the wall applies to her)
- - 69 N 69 N 69 m/s -69 m/s
To find the magnitude and direction of the average force the skater exerts on the wall, we can use Newton's second law of motion, which states that the force exerted on an object is equal to the mass of the object multiplied by its acceleration: F = m * a.
In this case, the skater is initially at rest and then moves backward with a velocity of -1.2 m/s. This change in velocity over the contact time of 0.8 seconds gives us the acceleration of the skater.
Acceleration (a) = Change in Velocity / Time
a = (-1.2 m/s - 0 m/s) / 0.8 s
a = -1.2 m/s / 0.8 s
a = -1.5 m/s²
Note: The negative sign indicates the direction of motion, which is opposite to the initial reference frame.
Now, we can substitute the given mass of 46 kg and the calculated acceleration into the equation to find the force exerted by the skater:
F = m * a
F = 46 kg * (-1.5 m/s²)
F = -69 N
The magnitude of the average force the skater exerts on the wall is 69 N, and the direction of the force is opposite to the initial direction of the skater's motion. Therefore, the correct answer is -69 N.