How long does it take for the mass to reach the bottom of the inclined place after it is released from rest?

Question

A mass is placed on a smooth inclined plane with an angle of 37 degrees wot the horizontal. If the inclined plane is 5.0-m long, how long does it take for the mass to reach the bottom of the inclined place after it is released from rest?

Answer 1

Force along plane = component of weight along plane = mg sin 37 deg = 0.6 mg

Acceleration a = 0.6 mg/m = 0.6 g = 0.6 * 9.8 m/s^2 = 5.88 m/s^2

Displacement s = 5 m

Initial velocity u = 0

s = ut + 1/2 at^2

5 = 0 + 1/2 * 5.88 * t^2

Or, t^2 = 5 * 2/5.88 = 1.7

t = sqrt(1.7) = 1.3 sec

Ans: 1.3 sec

Answer 2

acceleration = 9.8 sin 37 = 5.897787227 m/s^2

d= 0.5at^2

5= 0.5(5.897787227)t^2

t= 1.3 seconds

hope this helps  

Answer 3

Does smooth mean, the COF (coefficient of friction) is zero?