The Stuck Shopping Cart on a Ramp
A student pushes a 22 kg loaded cart up a 6.0 m long ramp that rises 0.90 m to a loading dock. She applies a steady horizontal push of 60 N along the ramp. A spring scale hooked to the cart reads a constant 18 N of friction while the cart rolls. Halfway up she stops pushing and the cart rolls back down 2.0 m before she catches it. The student claims the ramp is broken because the cart 'should not roll back if I pushed it that far.' A classmate says the cart must have been accelerating the whole way up.
1. Compute the gravitational potential energy gained per metre of ramp and the work done against friction over the full 6.0 m. Use g = 9.8 m/s².
Read the worked answer
Height gained per metre of ramp = 0.90 m / 6.0 m = 0.15 m per metre. Potential energy gain per metre = mgh = 22 kg × 9.8 m/s² × 0.15 m ≈ 32.3 J per metre. Over 6.0 m the total PE gain is 22 × 9.8 × 0.90 ≈ 194 J. Work against friction = friction force × distance = 18 N × 6.0 m = 108 J. Total work needed to move the cart steadily up the full ramp = 194 + 108 ≈ 302 J. The student's 60 N push over 6.0 m supplies 60 × 6.0 = 360 J, so a steady push is just enough with about 58 J to spare.
2. Evaluate the classmate's claim that the cart was accelerating the whole way up. Use the force balance.
Read the worked answer
Along the ramp, the component of weight opposing motion is mg sin θ. sin θ = 0.90/6.0 = 0.15, so that component is 22 × 9.8 × 0.15 ≈ 32.3 N. Adding 18 N of friction gives about 50.3 N of total resistance. The push is 60 N, so net force up the ramp is about 9.7 N and acceleration is 9.7/22 ≈ 0.44 m/s². So the classmate is partly right: with a constant 60 N push the cart does accelerate slowly upward. But the claim 'the whole way' fails once she stops pushing, because then the net force is 50.3 N down the ramp and the cart decelerates, stops, and rolls back - exactly what was observed.
3. The cart rolls back 2.0 m after she releases it. Explain why this is consistent with energy conservation, and state one assumption in your reasoning.
Read the worked answer
At release the cart has kinetic energy from the slow acceleration plus whatever height it had gained. Rolling back 2.0 m converts 22 × 9.8 × (2.0 × 0.15) ≈ 64.7 J of potential energy into kinetic energy, minus 18 × 2.0 = 36 J lost to friction, leaving about 28.7 J of kinetic energy at the catch point (plus the kinetic energy it had at release). Nothing is violated: the cart simply trades height for speed while friction removes some of the total. The key assumption is that friction stays a constant 18 N during the rollback; if the ramp surface or wheel alignment changed, the energy bookkeeping would shift, but the direction of the result would not.