Question

a man pushing a crate of mass m = 92.0 kg at a speed of v = 0.875 m/s encounters a rough horizontal surface of length l = 0.65 m as in the figure below. if the coefficient of kinetic friction between the crate and rough surface is 0.350 and he exerts a constant horizontal force of 285 n on the crate.
(a) find the magnitude and direction of the net force on the crate while it is on the rough surface.
magnitude n
direction - select -
(b) find the net work done on the crate while it is on the rough surface.
j
(c) find the speed of the crate when it reaches the end of the rough surface.
m/s

Explanation:

Step1: Calculate the frictional force

The normal force $N$ on the crate on a horizontal surface is equal to its weight, $N = mg$, where $m = 92.0$ kg and $g=9.8$ m/s². The frictional force $f_k=\mu_k N=\mu_kmg$.

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The applied force $F = 285$ N. The net - force $F_{net}=f_k - F$ (since the frictional force opposes the motion and is larger than the applied force).

$$F_{net}=315.56 - 285=30.56\text{ N}$$

The direction of the net - force is opposite to the direction of motion.

Step2: Calculate the net work done

The net work done $W_{net}=F_{net}d$, where $d = 0.65$ m.

$$W_{net}=30.56\times0.65 = 19.864\text{ J}$$

Step3: Use the work - energy theorem to find the final speed

The work - energy theorem states that $W_{net}=\Delta K=K_f - K_i$, where $K_i=\frac{1}{2}mv_i^2$ and $K_f=\frac{1}{2}mv_f^2$.

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Substitute $W_{net}=19.864$ J, $m = 92.0$ kg and $v_i = 0.875$ m/s.

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Answer:

(a) magnitude: $30.6$ N (rounded to one decimal place), direction: opposite to the direction of motion
(b) $19.9$ J (rounded to one decimal place)
(c) $1.09$ m/s