A cart is released on a frictionless 30° ramp.
The only force parallel to the ramp face is the weight's parallel component, mg sin θ.
Newton's second law turns that one force into the cart's acceleration, and something satisfying happens to the mass on the way.
On a frictionless incline, the net force parallel to the ramp face is the weight's parallel component: F_net = mg sin θ.
Apply a = F_net/m. So a = mg sin θ divided by m.
The mass cancels: a = g sin θ.
So every object on the same frictionless ramp has the same acceleration, whatever its mass.
For the 30° ramp, with sin 30° = 0.50: a = 9.8 × 0.50 = 4.9. The cart accelerates at 4.9 m/s² down the slope, and so would a cart ten times heavier.
A cart is released on a frictionless 30° ramp. Take g = 9.8 m/s² and sin 30° = 0.50. Calculate its acceleration.
Write down the values in the question:
g = 9.8 m/s²
θ = 30°
a = ?
Write down the equation:
a = g sin θ
Substitute in the values:
a = 9.8 × 0.50
Simplify:
a = 4.9 m/s²
The cart accelerates at 4.9 m/s² down the slope.
A skateboard is released on a frictionless 53° ramp. Take g = 9.8 m/s² and sin 53° = 0.80. Enter its acceleration in m/s².
A 2 kg cart and a 10 kg cart are released together on the same frictionless 30° ramp. Which statement about their accelerations is correct?
Steeper ramp, bigger sin θ, bigger acceleration. At 90° the ramp is a wall and a = g: free fall is the incline's limiting case.
A 5.0 kg tray is released on a frictionless 25° ramp. Take g = 9.8 m/s² and sin 25° = 0.42. Enter its acceleration in m/s².
A crate is released on a frictionless 37° chute. Take g = 9.8 m/s² and sin 37° = 0.60. Calculate its acceleration.
Write down the values in the question:
g = 9.8 m/s²
θ = 37°
a = ?
Write down the equation:
a = g sin θ
Substitute in the values:
a = 9.8 × 0.60
Simplify:
a = 5.88 m/s²
The crate accelerates at 5.88 m/s² down the slope, which is about 5.9 m/s².
A toolbox slides from rest down a frictionless 45° chute. Take g = 9.8 m/s² and sin 45° = 0.71. Enter its acceleration in m/s².
A delivery tote is released on a frictionless 15° conveyor ramp. Take g = 9.8 m/s² and sin 15° = 0.26. Enter its acceleration in m/s².
Casper's 8 kg sled and Lotte's 4 kg sled are released together on the same frictionless 20° slope. Which statement about their accelerations is correct?