A 50.0-kg block is being pulled up a 16.0° slope by a force of 250 N that is parallel to the slope. The coefficient of kinetic friction between the block and the slope is 0.200. What is the acceleration of the block?
A) 0.528 m/ $s ^ { 2 }$
B) 0.158 m/ $s ^ { 2 }$
C) 0.412 m/ $s ^ { 2 }$
D) 0.983 m/ $s ^ { 2 }$
E) 0.260 m/ $s ^ { 2 }$

Question

Quizplus · Accepted Answer

The force acting on the block can be resolved into two components, one parallel to the slope and one perpendicular to it. The perpendicular component, which is equal to $mg\cos{16.0°}$, balances the normal force, leaving only the parallel component, which is equal to $F_{parallel} = 250	ext{ N}$, to cause the block to move up the slope. The force of kinetic friction is equal to $F_f = \mu_kN$, where $N$ is the normal force and $\mu_k$ is the coefficient of kinetic friction. Therefore, the net force acting on the block is:
$F_{net} = F_{parallel} - F_f$
$= F_{parallel} - \mu_kN$
$= F_{parallel} - \mu_kmg\cos{16.0°}$
$= 250 - 0.200(50.0)(9.81)\cos{16.0°}$
$= 143.5	ext{ N}$
The acceleration of the block can now be found using Newton's second law:
$F_{net} = ma$
$a = \frac{F_{net}}{m}$
$= \frac{143.5}{50.0}$
$= 2.87	ext{ m/s}^2$
Since the block is moving up the slope, the acceleration is in the same direction as the parallel component of the applied force. Therefore, the answer is 0.412 m/s^2 (choice C).

A 500-Kg Block Is Being Pulled Up a 16 s2s ^ { 2 }s2

A 500-Kg Block Is Being Pulled Up a 16 $s ^ { 2 }$