A block is placed on a frictionless ramp at a height of 12.5 m above the ground.

Question

A block is placed on a frictionless ramp at a height of 12.5 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? 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?

Explanation / Answer

Since there is no friction, we can use conservation of potential and kinetic energy to determine the velocity at the bottom of the first ramp.

Initial P.E. = m * 9.8 * 12.5 = m * 122.5 J
K.E. = ½ * m * v2
Initial P.E. = Initial K.E.
m * 122.5 J = ½ * m * v2
v = 245
This is approximately 15.65 m/s.

Since there no friction of the second ramp, we can use conservation of potential and kinetic energy to determine the velocity at a height of 8.75 m.

Final P.E. = m * 9.8 * 8.75 = m * 85.75 J

To determine the block’s kinetic energy at this position, subtract the final potential energy from the initial potential energy.

KE = m * 122.5 J – m * 85.75 J = m * 36.75 J
½ * m * v2 = m * 36.75 J
v = 73.5
This is approximately 8.57 m/s.

As the block moved from the bottom of the second ramp to a height of 8.75 meters, its velocity decreased from approximately 15.65 m/s to 0 m/s at its highest point. As the block slid down the ramp, its velocity increased from 0 m/s to approximately 8.57 m/s.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.