a 2.0 kg block is 1.5 meters above the floor. the block slidesdown a ramp that i

Question

a 2.0 kg block is 1.5 meters above the floor. the block slidesdown a ramp that is inclined at 25 degree from the horizontal.assume that a constant 4.0 newton force of kinetic friction acts onthe second block as it slides down the ramp. a) how much work is done by friction as the block slides fromthe top of the ramp to the bottom of the ramp? b) how fast is the second block moving just before it reachesthe floor? a 2.0 kg block is 1.5 meters above the floor. the block slidesdown a ramp that is inclined at 25 degree from the horizontal.assume that a constant 4.0 newton force of kinetic friction acts onthe second block as it slides down the ramp. a) how much work is done by friction as the block slides fromthe top of the ramp to the bottom of the ramp? b) how fast is the second block moving just before it reachesthe floor?

Explanation / Answer

length of incline = height / sin25 = 1.5 / sin25 = 3.5493 meters . (a) work done by friction = - force *distance = - 4.0 * 3.5493 =    -14.197Joules . (b)   initial K + work by friction + work by gravity = final K .        0    -   14.197    +   m gh    =   (1/2) m v2 .              -14.197 + 2.0 * 9.8 * 1.5 = (1/2) * 2.0 *v2 .       v =   3.90 m/s    is its final speed .       v =   3.90 m/s    is its final speed

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