A 20 kilogram box 1/4 meter long is located at the center of a five meter long r

Question

A 20 kilogram box 1/4 meter long is located at the center of a five meter long ramp that is inclined at 45 degrees above horizontal. A pulley is at the ramp's upper end and a light weight string runs up and over the pulley connecting to a temporarily supported 5 kilogram mass below the pulley. The coefficient of static friction between the block and the ramp is 0.50 and the coefficient of kinetic friction is 0.40. a) If the support is removed from beneath the suspended 5 kilogram block, does the 20 kilogram block on the ramp slide, and if so, does it slide up or down the ramp? Show work to support your conclusion. b) If the block slips, how long does it take the "front" end of the block to reach the end of the ramp?

Explanation / Answer

When support is removed, Forces acting on 20 kg block are mg sin(45) along the incline downward.

mu(s) * N upward.

Calculate maximm friction = 0.5 * mg * cos 45 which is less than mgsin(45). So block will slide down. As the block slides down, kinetic friction coefficient acts.

So net force acting down will be mgsin(45) - mu(k) * mg cos45 =0.6* mg /sqrt(2) implies acceleration = 0.6g/sqrt(2)

Now s = 1/2 a t^2 so 2.5 + 0.25 = 0.3 g /sqrt (2) *t^2

t = sqrt(2.75 *sqrt(2) / 0.3g)

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