Final answer:
The force of friction exerted on the 64.5kg skater, who comes to rest from an initial speed of 2.40m/s in 3.52s, is approximately 43.86N.
Step-by-step explanation:
To calculate the force friction exerts on the 64.5kg skater coming to rest, we use Newton's second law which relates force (F), mass (m), and acceleration (a) by the formula F = m × a. First, we find the acceleration by using the change in velocity (Δv) over the time (t). The skater goes from an initial velocity of 2.40m/s to rest (0m/s), so Δv = -2.40m/s, and the time is 3.52s. The negative sign indicates a deceleration. The acceleration (a) is then Δv/t = (-2.40m/s) / 3.52s = about -0.68m/s². Using Newton's second law, the force of friction is F = m × a = (64.5 kg) × (-0.68 m/s²), which gives us approximately -43.86N. The negative sign indicates the force is in the opposite direction of motion, but since the question asks for the magnitude, we report 43.86N.