Question

A block slides down an inclined plane of slope angle theta with constant velocity. It is then projected up the same plane with an initial speed v0.

(a) How far up the incline will it move before coming to rest?
(b) Will it slide down again? Give an argument to back your answer.

Answer 1

a) S = v₀² / 4 g sin θ

Step-by-step explanation:

Let's apply Newton's second law, let's take a coordinate system with an axis parallel to the plane and the other perpendicular, in this case the only force that we have to decompose the weight (W)

Wx = W sin θ

Wy = W cos θ

First case. Body slides down

X axis

Wx-fr = 0

Axis y

N -Wy = 0

N = Wy

fr = Wx = W sint θ

Miu N = W sint θ

Miu W cos θ = Wsin θ

Miu = tan θ

Second case. Body raises the plane

X axis

Wx + Fr = m a

Axis y

N-Wy = 0

let's find the acceleration of the body going up

a = (Wx + fr) / m

fr = μ N = μ Wy

fr = μ mg cos θ

a = (mg sin θ + μ mg cos θ) / m

a = g (sin θ + μ cos θ)

a = g (sin θ + tan θ cos θ)

a = g (sin θ + sin θ)

a = g 2 sin 2

With the kinematic equation we find the distance that goes up, at the highest point the zero speed (vf = 0)

Vf² = v₀² - 2 a t S

0 = v₀² -2a S

S = v₀² / 2 a

S = v₀² / 2 (g 2sin θ)

S = v₀² / 4 g sin θ

b) in this case the block tries to slide down whereby the friction force opposes this movement

Wx- fr =, m a

mg sin θ - μ mg cos θ = m a

g (sin θ - μ cos θ) = a

a = g 2 sin θ

so that the body slides depends on the angle T for angles close to zero the body does not slide