A particle, initially at rest, slides down a frictionless ramp, of height h. Whe

Question

A particle, initially at rest, slides down a frictionless ramp, of height h. When it reaches the bottom of the ramp it slides a distance x along a table surface with friction (coefficient of kinetic friction mu_k), then slides off of the table and falls to the ground. The height of the table is also h. (You may give your answers in terms of g as well as any of the variables given in the question.) Find an expression for the speed of the particle at the bottom of the ramp. Find an expression for the speed of the particle as it just leaves the edge of the table. Find an expression for the speed of the particle an instant before it hits the ground? How far from the edge of the table does the particle land on the floor?

Explanation / Answer

(a) Initial energy of the particle = mgh
At bottom the energy will be kinetic only then = (1/2)mV2
Equating both
V = (2gh)1/2
(b) So at the bottom the velocity is V
and at the edge cosider its velocity is v
So chaneg in kinetic energy = work done
here work is done by friction
Force of friction = ukmg
Work done by friction = ukmg x
now
(1/2)mV2 - (1/2)mv2 = ukmg x
V2 - v2 = 2ukg x
v = (V2 - 2ukgx)1/2
(c) Vertical velocity when the particle leave the table would be zero.
Hence Vv2 = 2gh
horizontal component will remain same
Hence net velocity will be reultant of these two velocities.
Vground = (2gh + V2 -2ukgx)1/2
(d) Time taken by the particle to the ground from the table
V = u +at
(2gh)1/2 = 0 +gt
t = (2gh)1/2 /g
Therefore distance
= horizontal velocity*time
=  (V2 - 2ukgx)1/2 ( (2gh)1/2 /g )

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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