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A 2.6-kg block rests on a plane inclined at 35° to the horizontal. What is the frictional force acting on the block?

a) 44.2 N
b) 24.5 N
c) 16.8 N
d) 32.1 N

1 Answer

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Final answer:

In this case, we have a 2.6-kg block, so F_parallel = 2.6 kg * 10 m/s² * sin(35°) = 44.2 N.

Step-by-step explanation:

To calculate the frictional force acting on the block, we first need to determine the component of the weight force that is parallel to the incline.

We can use the formula F_parallel = m * g * sin(theta), where m is the mass of the block, g is the acceleration due to gravity (10 m/s²), and theta is the angle of the incline (35°).

To calculate the frictional force, use the formula F_parallel = m * g * sin(theta), where m is the mass of the block, g is the acceleration due to gravity, and theta is the angle of the incline.

Therefore, the frictional force acting on the block is 44.2 N. Therefore, the correct answer is (a) 44.2 N.

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