A block is placed on a frictionless ramp at a height of 14.5 m above the ground.

Question

A block is placed on a frictionless ramp at a height of 14.5 m above the ground. Starting from rest, the block slides down the ramp. At the bottom of the ramp, the block slides onto a frictionless horizontal track without slowing down. At the end of the horizontal track, the block slides smoothly onto a second frictionless ramp. How far along the second ramp does the block travel before coming to a momentary stop, as measured along the incline of the ramp? 14.5 m 46.3° 22.5 Number After the block comes to a complete stop on the second ramp, it will then begin moving back down the second ramp. What is the speed of the block when it is 8.75 m, vertically, above the ground? Number m/s

Explanation / Answer

(a)

Let, the second ramp is H height above the ground.

From law of energy conservation,

Total energy at starting position of block = Total energy at final postion at second ramp.

mgh = mgH

h = H = 14.5 m

Let, distance travel by block along second ramp = d

From pythagorus theorem,

sin(22.5) = H / d

d = 14.5 / sin(22.5)

d = 37.89 m

(b)

Let, speed of block at point 8.75 m above the ground.

Apply law of energy conservation-

TE at point 8.75 m above the ground = TE at final position at second ramp.

(1/2)mv^2 + mgh' = mgH

(1/2)*v^2 + 9.8*8.75 = 9.8*14.5

v = sqrt (2*9.8 (14.5 - 8.75))

v = 10.61 m/s

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