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
30.6
✓ n
direction
opposite as the motion of the crate✓
(b) find the net work done on the crate while it is on the rough surface.
×
the response you submitted has the wrong sign.
(c) find the speed of the crate when it reaches the end of the rough surface.
1.00
×
your response differs from the correct answer by more than 10%. double check your calculations. m/s

Explanation:

Step1: Calculate the frictional force

The normal force $N = mg$, where $m = 92.0$ kg and $g=9.8$ m/s². So $N=92.0\times9.8 = 901.6$ N. The frictional force $f=\mu_k N$, with $\mu_k = 0.350$. Then $f = 0.350\times901.6=315.56$ N. The applied force $F = 285$ N. The net - force $F_{net}=F - f=285 - 315.56=- 30.56\approx - 30.6$ N. The magnitude is $30.6$ N and the direction is opposite to the motion of the crate.

Step2: Calculate the net work done

The net work - energy theorem states that $W_{net}=F_{net}d$, where $d = 0.65$ m and $F_{net}=-30.6$ N. So $W_{net}=-30.6\times0.65=-19.89$ J.

Step3: Calculate the final speed

The initial kinetic energy is $K_i=\frac{1}{2}mv_i^2$, with $m = 92.0$ kg and $v_i = 0.875$ m/s, so $K_i=\frac{1}{2}\times92.0\times(0.875)^2=\frac{1}{2}\times92.0\times0.765625 = 35.215$ J. From the work - energy theorem $W_{net}=K_f - K_i$. We know $W_{net}=-19.89$ J and $K_i = 35.215$ J. Then $K_f=W_{net}+K_i=-19.89 + 35.215 = 15.325$ J. Since $K_f=\frac{1}{2}mv_f^2$, we have $v_f=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2\times15.325}{92.0}}=\sqrt{\frac{30.65}{92.0}}\approx0.577$ m/s.

Answer:

(b) - 19.89 J
(c) 0.577 m/s