Question

An object with a mass m = 3.5 kg is released from rest at the top of the ramp. The length of the ramp is 4 m. The object slides down the ramp reaching a speed of 1.8 m/s at the bottom.

(a) How much time (in sec) does it take the object to reach the bottom of the ramp? (use kinematics equations)

(b) What is the acceleration of the object (in m/s2 )? (use kinematics equations)

(c) If the force of kinetic friction between the ramp and the object is 20 N find the angle between the ramp and the horizontal θ (in degrees).

(d) What speed (in m/s) does the object reach at the bottom if the ramp is frictionless? Assume that the angle is the same as your answer for part (c).

(e) How much time (in sec) does it take the object to reach the bottom of the ramp in part (d)?

(f) What is the acceleration of the object (in m/s2 ) in part (d).

Answer 1

Given,

m = 3.5 kg ; L = 4 m ; vf = 1.8 m/s

a)v^2 = u^2 + 2 a s

1.8^2 = 0^2 + 2 a 4

a = 1.8^2/(8) = 0.405 m/s^2

v = u + at

t = v/a = 1.8/.405 = 4.44 s

Hence, t = 4.44 s

b)a = 0.405 m/s^2

(calculated in a)

c)Ff = 20

from Newtons second law

Fnet = ma

m a = m g sin(theta) - uk m g cos(theta)

3.5 x 0.405 = 3.5 x 9.8 sin(theta) - 20

sin(theta) = 0.6244

theta = sin^-1(0.6244)

theta = 38.64 deg

d)for frictionless case

mgh = 1/2 m v^2

v = sqrt (2 g h)

h = 4 sin38.64 = 2.5 m

v = sqrt (2 x 9.8 x 2.5) = 7 m/s

Hence, v = 7 m/s

e)v = u + at

7 = 0 + 6.12 t (a calculated in next part)

t = 7/6.12 = 1.14 s

Hence, t = 1.14 s

f)a = g sin(theta)

a = 9.8 x sin38.64 = 6.12 m/s^2

Hence, a = 6.12 m/s^2