A block of mass 70 kg is placed on an incline plane whose angle is 24 degrees. t

Question

A block of mass 70 kg is placed on an incline plane whose angle is 24 degrees. the coefficient of friction is 0.15. the block slides 3 meters down the incline to the bottom of the plane. what is its speed at the bottom of the plane? how much work was done by friction as it moves to the bottom of the plane? what is the total work done as the block moves to the bottom of the plane? after reaching the bottom of the plane the block slides 2.75 meters and stops, what is the coefficient of kinetic friction between the block and the level surface?

Explanation / Answer

Energy conservation:
mgh - mgcos24.s =(1/2) mv2

s=3

h = 3 sin24

g=9.8

v = 3.98m/s...Ans

Work done by friction =mgcos24.s = 282J...Ans

Total work done = change in KE = (1/2) mv2 =554.4 J...Ans

mg *2.75 =(1/2) mv2

= 0.294 ...Ans

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