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A block of mass 15.0 kg slides down a ramp inclined at 28.0 ∘ above the horizontal. as it slides, a kinetic friction force of 30.0 n parallel to the ramp acts on it. if the block slides for 7.00 m along the ramp, find the work done on the block by friction.

A) 150 J150J
B) 188 J188J
C) 200 J200J
D) 220 J220J

1 Answer

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Final answer:

The work done on the block by friction as it slides down a ramp with a kinetic friction force of 30.0 N over a distance of 7.00 m is -210 J in magnitude.

Step-by-step explanation:

The work done by friction on the block can be calculated using the formula for work, which is work = force × displacement × cos(θ). In this case, the force of friction is parallel to the displacement and opposite to the motion, which means θ=180 degrees or π radians, so the cosine term is -1. Therefore, to answer the question:

Work done by friction = 30.0 N × 7.00 m × cos(180°) = 30.0 N × 7.00 m × (-1) = -210 J

Since work done against the direction of motion is considered negative, the magnitude of the work done is 210 J, technically not matching any of the answers given 'A' through 'D'. It's possible there might have been a mistake in the question or the answer choices. The correct closest answer should have been 'B' 210 J if it were an option.

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