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 calculate the weight of the block that is acting perpendicular to the incline:

$W_{\perp}= mg\cos{	heta} = (4	ext{ kg})(9.81	ext{ m/s}^2)\cos{30^{\circ}} = 33.91	ext{ N}$

Next, we need to calculate the maximum force of static friction that is acting parallel to the incline:

$F_f= \mu_s W_{\perp} = (0.700)(33.91	ext{ N}) = 23.74	ext{ N}$

Finally, we need to find the minimum force required to start the block moving up the incline. This force must be greater than the force of static friction:

$F > F_f = 23.74	ext{ N}$

The only answer choice that satisfies this condition is D) 84.0 N.

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