Well, let's see how far down it went:
sin 30 = h/3
so
h = 1.5 meters high
so at the top it has potential energy = m g h = 10*9.81*1.5 = 147 Joules compared to when it hits the ground.
-------------------------------------
Let's look at the friction:
weight = m g = 98.1 N
so
normal force on ramp = 98.1 cos 30
= 85 N
so the friction force = .35*85
= 29.7 N
The component of weight down the ramp is
98.1 sin 30 = 49.1 N
-------------------------
The net force down the ramp is thus
49.1 - 29.7 = 19.4 N (part (1))
------------------------
The Ke at bottom is the loss in potential energy - work done by friction
= 147 Joules - 29.7*3 = 147-89.1
= 57.9 Joules
Alternately the work done by the net force down the ramp is
19.4 N
so the work done is 19.4*3 = 58.2 Joules
so call it 58 Joules
--------------------------------------
(1/2) m v^2 = 58
v^2 = 58 * 2 /10
v = 3.4 m/s
Starting from rest, a 10kg suitcase slides down a 3.00m long ramp inclined at 30 degrees from the floor. The coefficient of friction between the suitcase and the ramp is 0.35.
1) what net force is applied to the suitcase while it is on the ramp?
2)what is the change in the kinetic energy of the suitcase as it slides down the ramp?
3) how fast is it traveling when it reaches the bottom of the ramp?
1 answer
3 answers