Question

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?

Answer 1

Answer:

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