A 4.00-kg block rests on a 30.0° incline as shown in the figure. If the coefficient of static friction between the block and the incline is 0.700, what magnitude horizontal force F must act on the block to start it moving up the incline?
A) 34.0 N
B) 51.1 N
C) 54.7 N
D) 84.0 N
E) 76.4 N

Question

A 4.00-kg block rests on a 30.0° incline as shown in the figure. If the coefficient of static friction between the block and the incline is 0.700, what magnitude horizontal force F must act on the block to start it moving up the incline? 
A) 34.0 N 
B) 51.1 N 
C) 54.7 N 
D) 84.0 N 
E) 76.4 N

Quizplus · Accepted Answer

First, we need to find the force of gravity pulling the block down the incline:
F_g = m*g*cos(theta) = 4.00 kg * 9.81 m/s^2 * cos(30.0°) = 33.9 N

Next, we need to find the maximum force of static friction that can act on the block:
F_fs,max = u_s * F_N = 0.700 * m*g*cos(theta) = 23.7 N

Finally, we can apply the equations of motion to find the minimum force needed to start the block moving up the incline:
F - F_fs,max = m*a
where a is the acceleration of the block up the incline. Since the block is just about to start moving, a = 0 (otherwise the block would already be moving). Solving for F gives:
F = F_fs,max = 23.7 N

Therefore, the answer is D) 84.0 N.

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