Question

A body with mass m slides down a frictionless ramp inclined at 600, with an initial speed v1 = 3 m/s,

Answer 1

Going down the first ramp:

• net force parallel to the ramp:

∑ F = W sin(60°) = m a₁

(W for weight)

• net force perpendicular to the ramp:

∑ F = N + W cos(60°) = 0

(N for normal force)

The body has mass 0.1 kg, and with g = 10 m/s², its weight is W = 1 N. So in the first equation, we get

(1 N) sin(60°) = (0.1 kg) a₁   →   a₁ ≈ 8.7 m/s²

Let d₁ be the distance the body moves down the ramp, i.e. the distance along the ramp between the starting point and the point O. Then

sin(60°) = h / d₁   →   d₁ = 2h/√(3) ≈ 1.15h

Given an initial speed v₁ = 3 m/s, we find the speed v₂ at point O to be

v₂² - (3 m/s)² = 2 (8.7 m/s²) d₁

v₂ ≈ √(9 m²/s² + (17.4 m/s²) (1.15h))

v₂ ≈ √(9 m²/s² + (20 m/s²) h)

Going up the second ramp:

• net parallel force:

∑ F = -Fr - W sin(30°) = m a₂

(Fr for friction)

• net perpendicular force:

∑ F = N - W cos(30°) = 0

sin(30°) = cos(60°), and cos(30°) = sin(60°), so the second equation gives N = 0.87 N. Then with µ = 0.1, we have Fr = µ N = 0.087 N. The first equation then gives

-0.087 N - 0.5 N = (0.1 kg) a₂   →   a₂ ≈ -5.9 m/s²

We now have

tan(30°) = h/R   →   h = (2.5 m)/√3 ≈ 1.4 m

(which, by the way, tells us that v₂ ≈ 6.2 m/s)

Then the distance traveled up the ramp is

d₂ = √(h² + R ²) ≈ 2.9 m

Use this to solve for the speed at the top of the ramp:

v₃² - v₂² = 2 (-5.9 m/s²) d₂

v₃ = √((6.2 m/s)² - (11.8 m/s²) (2.9 m)) ≈ 2.0 m/s

At the top of the second ramp:

• net parallel force:

∑ F = -Fsp - W sin(30°) = m a₂

(Fsp for spring)

• net perpendicular force:

∑ F = N - W cos(30°) = 0

By Hooke's law, Fsp = kx, so in the first equation we get

-k (0.10 m) - (1 N) cos(60°) = (0.1 kg) (-5.9 m/s²)

→   k ≈ 0.87 N/m