A box is released from rest at the top of a 30 degree ramp. The box slides down the ramp, dropping a vertical distance of 1.5 m to the floor. How long does it take to reach the floor? Neglect friction.
The ramp is 3.0 m long. If we knew the acceleration of the box we could use the constant acceleration equations to find the time.
By analyzing the forces on the box the acceleration can be determined. Once again, the key is to construct a good free-body diagram.
Which way will the box accelerate? With no friction, the box will accelerate down the ramp.
Align the coordinate system with the acceleration.
+x = down the ramp and +y = perpendicular to the ramp.
The forces need to be split into components. The normal force is OK - it's in the +y direction. Split the force of gravity into a component down the ramp, and a component perpendicular to the ramp.
mg_{x} = mg sin(q) mg_{y} = mg cos(q) |
SF_{x} = m a_{x}
mg sin(q) = m a_{x}
a_{x} = g sin(q)
Plugging in g = 9.8 and q = 30 degrees:
a_{x} = 4.9 m/s^{2}
Plug this, and x_{o} = 0, v_{ox} = 0, and x = 3 m, into the equation:
x - x_{o} = v_{ox} t + ½ a_{x} t^{2}
gives:
3 = 2.45 t^{2}
solving for t tells us that the box takes 1.11 s to slide down the ramp.