Question

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correct. the block is at rest and, therefore, in equilibrium. according to newtons second law, then, the net force acting on the block in a direction parallel to the inclined surface of the incline must be zero. this means that the force of static friction directed up the incline must balance the component of the blocks weight directed down the incline (8.0 kg)(9.8 m/s²) sin 22° = 29 n.
the drawing shows a block at rest on an incline. the mass of the block is 8.0 kg. what is the static frictional force that acts on the block?
73 n
32 n
78 n
29 n
there is not enough information to calculate the frictional force.

Explanation:

Step1: Identify forces acting on block

The block is at rest on the incline, so net - force in the direction parallel to the incline is zero. The two forces acting parallel to the incline are the component of the weight of the block down the incline and the static - friction force up the incline.

Step2: Calculate the component of the weight down the incline

The formula for the component of the weight of an object of mass $m$ on an incline of angle $\theta$ is $F = mg\sin\theta$, where $g = 9.8\ m/s^{2}$, $m = 8.0\ kg$ and $\theta=22^{\circ}$. So $F=(8.0\ kg)\times(9.8\ m/s^{2})\times\sin22^{\circ}$.
Using a calculator, $\sin22^{\circ}\approx0.375$, then $F = 8.0\times9.8\times0.375=29.4\ N\approx29\ N$.
Since the block is in equilibrium, the static - friction force $f_s$ balances this component of the weight. So $f_s = 29\ N$.

Answer:

D. 29 N