Question

# 3. A force of 70N is applied to a 10kg box with a friction force of 30N opposing it's motion. The box starts from rest and reaches a speed of 5m/s. Determine the distance over which the force acts on the box.

Answer

7.81m

- Q: What is the acceleration of the box? A: The acceleration of the box can be found using the formula a = (F - f) / m, where F is the force applied, f is the friction force, and m is the mass of the box. a = (70N - 30N) / 10kg = 4m/s^2.
- Q: How long does it take for the box to reach the speed of 5m/s? A: The time it takes for the box to reach a speed of 5m/s can be found using the formula v = at, where v is the final velocity, a is the acceleration, and t is the time. 5m/s = 4m/s^2 * t, t = 1.25s.
- Q: What is the distance over which the force acts on the box? A: The distance can be found using the formula d = vt + 1/2at^2, where d is the distance, v is the final velocity, a is the acceleration, and t is the time. d = 5m/s * 1.25s + 1/2 * 4m/s^2 * (1.25s)^2 = 7.81m.