Question

A 50 kg crate slides down a ramp that is inclined by 30 degrees from the horizontal. The acceleration of the crate parallel to the surface of the ramp is 2.0m/s^2, and the length of the ramp is 10m.

Determine the kinetic energy accumulated by the crate when it reaches the bottom of the ramp if it started from rest at the top of the incline.

Calculate the amount of energy lost by the crate due to friction in its journey down the ramp.

Compute the magnitude of the frictional force that acts on the crate as it slides down the ramp and calculate the coefficient of kinetic friction between the ramp and the crate.

Answer 1

given that :

mass of the crate, m = 50 kg

making an angle from the horizontal, = 30 degree

acceleration of the crate, a = 2 m/s²

length of the ramp, h = 10 m

(a) the kinetic energy accumulated by the crate when it reaches the bottom of the ramp if it started from rest at the top of the incline which is given as ::

using a conservation of energy,

kinetic energy = potential energy

1/2 mv² = mah

v² = 2 a h                  

v = 2 a h                                { eq. 1 }

inserting the values in eq.1

v = 2 (2 m/s²) (10 m)

v = 40 m/s

v = 6.32 m/s

kinetic energy is given as, K.E = 1/2 m v²                            { eq. 2 }

K.E = (0.5) (50 kg) (40 m/s)

K.E = 1000 J

(b) the amount of energy lost by the crate due to friction in its journey down the ramp which is given as :

if friction was not present, K.E = (0.5) (50 kg) (9.8 m/s²) (10 m) = 2450 J

energy lost by crate, K.E_loss = K.E_friction - K.E_total                        { eq. 3 }

K.E_loss = (2450 J) - (1000 J)

K.E_loss = 1450 J

(c) the coefficient of kinetic friction between the ramp and the crate which is given as :

from above figure,

F_gx = mg sin                     { eq. 4 }

and f_k = kinetic frictional force = _k mg cos                           { eq. 5 }

F_net = F_gx - f_k

where, F_net = m a

m a = mg sin - _k mg cos                   { eq. 6 }

a = g sin - _k g cos

(2 m/s²) = (9.8 m/s²) sin 30⁰ - _k (9.8 m/s²) cos 30⁰

(2 m/s²) = (4.9 m/s²) - _k (8.48 m/s²)

(2.9 m/s²) = (8.48 m/s²) _k

_k = 0.341