A 1.8-kg block is released from rest at the top of a rough 30$\degree$inclined plane. As the block slides down the incline, its acceleration is 3.0 m/s2 down the incline. Determine the magnitude of the force of friction acting on the block.
A)4.2 N
B)3.0 N
C)3.4 N
D)3.8 N
E)2.3 N

Question

Quizplus · Accepted Answer

We can use the equation $F_{net}=ma$ to solve for the force of friction. The net force in the direction of motion down the incline is given by $F_{net}=mg\sin	heta-F_f$, where $mg\sin	heta$ is the force component down the incline due to gravity and $F_f$ is the force of friction. Substituting the given values, we get:
$$F_{net}=ma$$
$$mg\sin	heta-F_f=ma$$
$$(1.8\ kg)(9.8\ m/s^2)(\sin 30^\circ)-F_f=(1.8\ kg)(3.0\ m/s^2)$$
$$F_f=34.3\ N-(5.4\ N)$$
$$F_f=28.9\ N \approx \boxed{3.4\ N}$$

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