A 2 kg block is at the top of a 5m tall ramp that has an angleof 30 degrees. The

Question

A 2 kg block is at the top of a 5m tall ramp that has an angleof 30 degrees. The block is allowed to slide down. Imagine that theramp has a coefficient of friction of 0.2. What is the work done byfriction during the 2 kg block's slide? The correct answer is 34 N, but I do not understand how to getthat answer. A 2 kg block is at the top of a 5m tall ramp that has an angleof 30 degrees. The block is allowed to slide down. Imagine that theramp has a coefficient of friction of 0.2. What is the work done byfriction during the 2 kg block's slide? The correct answer is 34 N, but I do not understand how to getthat answer.

Explanation / Answer

the distance of the ramp from top to bottom is s and: sin(30) = 5m/s => s = 5m / sin30 = 10m the friction force is: F=mgcos30 =2*9.8m/s2*cos30*0.2 = 3.39N the work done W = F*s = 3.39N*10m = 33.9J ˜ 34J it can't be 34N since work done must be in J. .

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