A box is released from rest at the top of a 30 degree ramp of length 3 m. The box slides down the ramp, dropping a vertical distance of 1.5 m to the floor. Neglect friction. How long does it take the box to reach the floor?
Steps 1-3: See above diagram
Step 4: Take components
|max = mg sin(θ)|
may = mg cos(θ)
Step 5: Apply Newton's 2nd law
ΣFx = m ax gives mg sin(θ) = m ax
Step 6: Solve
ax = g sin(θ)
Plugging in g = 9.8 and θ = 30 degrees gives: ax = 4.9 m/s2
Step 7: Check! &theta=0;
gives ax=0; &theta=90o gives ax=g.
Finally: How long does it take the box to reach the floor?
Plug ax = 4.9 m/s2, x(0) = 0, vx(0) = 0, and x = 3 m, into the kinematic equation:
x(t) = x(0) + vx(0) t + ½ ax t2
This gives 3 = 2.45 t2
Solving for t gives 1.11 s as the time for the box to slide down the ramp.