CP An 8.00-kg block of ice, released from rest at the top of a 1.50-m-long frictionless ramp, slides downhill, reaching a speed of 2.50 m/s at the bottom. (a) What is the angle between the ramp and the horizontal? (b) What would be the speed of the ice at the bottom if the motion were opposed by a constant friction force of 10.0 N parallel to the surface of the ramp?
Solution 16E Step 1: In this question, we need to find The angle between the ramp and horizontal Speed of the ice at the bottom of the inclined plane Considering the data given Mass of ice block m = 8 kg Initial position on ramp d = 1.50 m Final velocity v =f2.50 m/s Step 2 : We shall find the acceleration down the ramp Let us consider the equation vf2 = vi 2+ 2ad (f vi) a1= 2d Substituting the values we get (2.5 m/s 0 m/s ) a1= 2×1.5m a = 2.083 m/s 2 1 Hence we have acceleration down the slope as 2.083 m/s 2 Step 2: With this acceleration, we shall find the angle made by the horizontal with inclined plain This is obtained using relation a = g × sin a sin = g Substituting the values we get sin = 2.083 m/s 9.8 m/s 1 = sin ( 0.2125) = 12. 27 0 0 This can be approximated to 12 0 Hence we have the angle made by the horizontal with inclined plain as 12