What is the angle of the ramp and speed of the block of ice in the following case:
A 8.20 kg block of ice, released from rest at the top of a 1.10m long frictionless ramp, slides downhill, reaching a speed of 2.99m/s at the bottom. What is the angle between the ramp and the horizontal?
What would be the speed of the ice at the bottom if the motion were opposed by a constant friction force of 10.2N parallel to the surface of the ramp?
There is a block of ice weighing 8.2 kg placed on the ramp. Let the angle between the ramp and the horizontal be X. The gravitational force of attraction between the Earth and the block accelerates it downwards with an acceleration equal to 9.8 m/s^2 acting in a vertically downwards direction. This can be divided into two component, one parallel to the ramp and acting downwards and the other perpendicular to the ramp. The speed acquired by the block as it slides down is due to the parallel component that is equal to 9.8*sin X.
The block after sliding 1.1 m has a velocity of 2.99 m/s, 2.99^2 = 2*a*1.1 where a is the acceleration
=> a = 2.99^2/2.2
9.8*sin X = 2.99^2/2.2
=> X = arc sin (2.99^2/21.56)
=> X = 24.49 degrees
If the motion of the ice were opposed by a parallel force of 10.2 N to the surface of the ramp there is a net acceleration of 9.8*sin 24.49 - 10.2/8.2 = 2.818 m/s^2.
As it slides down 1.1 m, v^2 = 2*2.818*1.1
=> v = 2.48 m/s
The speed of the block at the bottom of the ramp would be 2.48 m/s.