An object with a mass of 10.0 kg is at rest at the top of a frictionless incline
ID: 1488931 • Letter: A
Question
An object with a mass of 10.0 kg is at rest at the top of a frictionless inclined plane of height 8.00 m and an angle of inclination 30.0 degree with the horizontal. The object is released from this position and it stops at a distance d from the bottom of the inclined plane along a horizontal surface, as shown in Figure 8-9. The coefficient of kinetic friction for the horizontal surface is 0.400 and g = 10.0 m/s^2. Refer to Figure 8-9. What is the speed of the object at the bottom of the inclinedExplanation / Answer
answer is E)10.0m/s
Vertical height = 8 sin 30 = 4 m
Velocity at the bottom of the plane u^2 = 2gh
Along the horizontal surface the acceleration = g
If d is the distance traveled then u^2 = 2 ad = 2 g d
2 g d = 2gh
d = h
d =h/ = 4/0.4 = 10 m
For your additional information and knowledge:
The problem can be answered by simple analytical method. as followed.
In the vertical direction there is no friction.
If there were no friction, in the horizontal direcion also, then it will never stop and move with the same velocity for ever.
If in the horizontal direction if the acceleration is again "-g" it will move through the distance equal to the vertical hieght namely 4 m.
Actual deceleraion is 0.4 times g. Hence the horizontal distance will increse by an amount of 1/0.4 of the vertical height, i.e, 4/0.4 = 10 m.