Question

A body with mass m slides down a frictionless ramp inclined at 600, with an initial speed v1 = 3 m/s,
starting at height h, and continues its motion upward on a 300 inclined ramp with a friction. Than this
body hits a spring which is fixed to a wall and compresses it 10 cm. The spring is initially at a distance R
on horizontal plane, and at height h. Find:
a) The speed v2 at the point 0, where the body starts to climb up to 300 inclined ramp, in terms of h.
b) The speed v3 where it hits the spring.
c) The spring constant k.
m=0.1 kg, µk=0.1, R=2.5 m., v1=3 m/s, Cos(600) = 0.5, Sin(600)=0.87, x=10 cm. and g= 10 m/s2 .

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