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Solve for the force of friction acting on a 50-kg carton sliding at 4 m/s if it stops in 3 seconds.

a) 23 N
b) 600 N
c) 67 N
d) 45 N

Answer :

Final answer:

The force of friction acting on a 50-kg carton to bring it to a stop in 3 seconds is 67 N, by calculating the acceleration needed to stop the carton and then applying Newton's second law.

Option c.

Explanation:

To solve for the force of friction acting on a 50-kg carton sliding at 4 m/s that stops in 3 seconds, we will first determine the acceleration required to bring the carton to a stop.

Since the initial velocity (v_i) is 4 m/s and the final velocity (v_f) is 0 m/s (because the carton stops), and the time (t) it takes to stop is 3 s, we can use the formula for acceleration (a):

a = (v_f - v_i) / t = (0 - 4) m/s / 3 s = -4/3 m/s².

The acceleration is negative because it is a deceleration. Next, we use Newton's second law, F = ma, to find the force (F) responsible for this deceleration. We have:

F = ma = (50 kg)(-4/3 m/s²) = -200/3 N.

The negative sign indicates that the force is in the opposite direction of movement. Since this is the force required to stop the carton, it is the same in magnitude as the frictional force, but in the opposite direction. Thus, the frictional force (f_friction) is:

f_friction = |F| = 200/3 N ≈ 66.67 N, which rounds to 67 N, answer choice (c).

Option c.

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