A 1.0 kg block slides along a frictionless horizontal surface with a speed of 7.0 m/s. After sliding a distance of 2.0 m, the block makes a smooth transition to a frictionless ramp inclined at an angle of 40° to the horizontal.

How far up the ramp does the block slide before coming momentarily to rest?

3.9 m

Explanation:

The principle of work and energy

ΔE = W Formula (1)

where:

ΔE: mechanical energy change (J)

W : work of the non-conservative forces (J)

ΔE = Ef - E₀

Ef : final mechanical energy

E₀ : initial mechanical energy

Ef = K f + Uf

E₀ = K₀ + U₀

K = (1/2) mv² : Kinetic energy (J)

U = mgh : Potential energy (J)

m: mass (kg)

v : speed (m/s)

h: hight (m)

Known data

m = 1 kg : mass of the block

v₀ = mg (h).0 m/s Initial speed of the block

vf = 0 = Final speed of the block

θ = 40° : angle θ of the ramp with respect to the horizontal direction

μk=0 : coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

Problem development

W = 0, Because the friction force (non-conservative force) is zero

Principle of work and energy to the Block:

ΔE = W

Ef - E₀ = 0 Equation (1)

Ef = K f + Uf = (1/2) m (0) ² + mg (h) = mg (h) (Joules)

E₀ = K₀ + U₀ = (1/2) m (7) ² + mg (0) = 24.5m (Joules)

In the equation (1):

Ef = E₀

mg (h) = 24.5m

We divided by m both sides of the equation

g (h) = 24.5

h = 24.5 / g

h = 24.5 / 9.8

h = 2.5 m

We apply trigonometry at the ramp to calculate how far up the ramp (d) does the block slide before coming momentarily to rest:

sinθ = h/d

d = h / sinθ

d = 2.5 m / sin40°

d = 3.9 m