Question

A 10-lb block has a speed of 4 ft/s when a force of f=(8t²) lb is applied. The coefficient of kinetic friction at the surface is μk = 0.2. What is the value of t when the block comes to rest?

1) t = 0.5 s
2) t = 1 s
3) t = 1.5 s
4) t = 2 s

Answer 1

The block comes to rest when t = 1 s.

Step-by-step explanation:

The equation to calculate the force of friction is µkN, where µk is the coefficient of kinetic friction and N is the normal force.

In this problem, the normal force is equal to the weight of the block, which is 10 lbs or 10 lb × 32.2 ft/s² (since 1 lb is approximately equal to 32.2 ft/s² in Imperial units) = 322 ft lb/s².

Given that the coefficient of kinetic friction µk = 0.2, we can calculate the force of friction:

1. Friction force = 0.2 × 322 ft lb/s² = 64.4 lb ft/s²

At the point where the block comes to rest, the force of friction is equal and opposite to the applied force:

1. F = 64.4 lb ft/s²
2. 8t² = 64.4 lb ft/s²
3. t² = 8 lb ft/s² / 8 lb = 1 s²
4. t = √(1 s²) = 1 s

Therefore, the correct answer is t = 1 s (option 2).