Spring-Launched Cart on a Frictional Incline
A 2.0 kg laboratory cart is attached to a spring with k = 800 N/m on a horizontal track. The spring is compressed 0.30 m from equilibrium and released. The cart leaves the spring at the equilibrium point and then travels 0.50 m along the horizontal track before entering a 30.0° incline. Friction is negligible while the cart is in contact with the spring, but the coefficient of kinetic friction is 0.25 on the horizontal stretch and on the incline. Air resistance is negligible, and the cart is treated as a point mass. A photogate at the bottom of the incline measures the cart's speed as it starts up. The group predicts the cart will travel 2.0 m up the incline. They also plan a second trial with the spring compressed 0.60 m and expect exactly twice the distance. The apparatus has a maximum incline length of 4.0 m and a bumper at the top.
1. Calculate the cart's speed at the bottom of the incline and the distance it travels up the incline. Compare with the group's 2.0 m prediction.
Read the worked answer
Spring potential energy = 0.5kx^2 = 0.5 × 800 × (0.30)^2 = 36 J. On the 0.50 m horizontal stretch, friction does Wf = μmgd = 0.25 × 2.0 × 9.8 × 0.50 = 2.45 J. Energy at the base of the incline = 36 − 2.45 = 33.55 J.
Speed at the base: v = sqrt(2E/m) = sqrt(2 × 33.55/2.0) = 5.79 m/s.
On the incline, energy is spent against gravity and friction: mgd sin30 + μmg cos30 d = d(2.0 × 9.8 × 0.5 + 0.25 × 2.0 × 9.8 × 0.866) = d(9.8 + 4.24) = 14.04d. Set 14.04d = 33.55, so d = 2.39 m. The group's 2.0 m prediction is too short by about 0.4 m.
2. Evaluate the prediction that doubling the compression will exactly double the distance traveled up the incline.
Read the worked answer
Doubling the compression from 0.30 m to 0.60 m quadruples the spring energy, because E ∝ x^2: 0.5 × 800 × 0.36 = 144 J. Subtract the fixed 2.45 J horizontal friction loss: 141.55 J at the incline base.
Then d = 141.55/14.04 = 10.1 m. That is more than double the 2.39 m result and exceeds the 4.0 m track, so the cart hits the bumper before stopping. The prediction of exactly twice the distance is wrong because the spring energy scales quadratically while the horizontal friction loss is constant.
3. For the 0.60 m trial, calculate the cart's speed at the 4.0 m bumper and state whether it will tend to slide back after the bumper releases it.
Read the worked answer
At 4.0 m up the incline, energy left = 141.55 − 14.04 × 4.0 = 141.55 − 56.16 = 85.39 J. Speed = sqrt(2 × 85.39/2.0) = sqrt(85.39) = 9.24 m/s, so the bumper must absorb a large impact.
If the bumper releases the cart, the downhill gravity component is mg sin30 = 9.8 N. If the coefficient of static friction is about 0.30, maximum static friction is 0.30 × mg cos30 = 0.30 × 2.0 × 9.8 × 0.866 = 5.09 N. Because 9.8 N > 5.09 N, gravity exceeds static friction, so the cart will slide back down after release rather than staying at the top.