Question

A 5 kg box is tossed across the floor at 4 m/s and slides to a stop in 3 s. What is the average force of friction?

Answer 1

Answer on Question 48991, Physics, Mechanics | Kinematics | Dynamics

Question:

A 5 kg box is tossed across the floor at 4 m/s and slides to a stop in 3 s. What is the average force of friction?

Solution:

From the definition of the impulse we have:

$$\bar{F} \Delta t = m v,$$

So, substituting data from the conditions of the problem we obtain:

$$\bar{F} = \frac{m v}{\Delta t} = \frac{5 k g \cdot 4 \frac{m}{s}}{3 s} = 6.7 N.$$

Answer:

Average force of friction is 6.7N.

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