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
ma_{x} = mg sin(θ) ma_{y} = mg cos(θ) |
Step 5: Apply Newton's 2nd law
ΣF_{x} = m a_{x} gives mg sin(θ) = m a_{x}
Step 6: Solve
a_{x} = g sin(θ)
Plugging in g = 9.8 and θ = 30 degrees gives: a_{x} = 4.9 m/s^{2}
Step 7: Check! &theta=0;
gives a_{x}=0; &theta=90^{o} gives a_{x}=g.
Finally: How long does it take the box to reach the floor?
Plug a_{x} = 4.9 m/s^{2}, x(0) = 0, v_{x}(0) = 0, and x = 3 m, into the kinematic equation:
x(t) = x(0) + v_{x}(0) t + ½ a_{x} t^{2}
This gives 3 = 2.45 t^{2}
Solving for t gives 1.11 s as the time for the box to slide down the ramp.