Question

Given: m = 84 kg & v = 6.7 m/s. Find:

a) Magnitude of impulse (I)
b) Average friction force, if t = 0.75 seconds
a) 563.8 N∙s, 75.17 N
b) 4510.4 N∙s, 601.39 N
c) 3175.8 N∙s, 423.44 N
d) 2255.2 N∙s, 300.69 N

Answer 1

The magnitude of the impulse is 563.8 N·s, and the average friction force is 751.73 N. The impulse is calculated by multiplying the mass by the velocity, assuming a final velocity of 0 m/s.

Step-by-step explanation:

The student's question pertains to finding the magnitude of the impulse (I) and the average friction force given the mass (m), velocity (v), and time (t). The formula to calculate impulse is I = m × Δv, where Δv is the change in velocity. However, to find the Δv, we would need the final velocity, which is not provided, so we assume that the object comes to a rest (final velocity = 0 m/s), making Δv equal to the initial speed v. Therefore, I = m × v = 84 kg × 6.7 m/s = 563.8 N·s. To find the average friction force, we use the formula impulse equals average force times time (Δt), so F = I / t = 563.8 N·s / 0.75 s = 751.73 N. The correct answer would be a).