How much energy reaches the spring?
One slider on one continuous route: accelerate downhill, lose energy to friction, then compress a spring. Not three separate experiments.
The problem
Release a 1 kg point slider from rest at the top of a smooth 3 m incline at 30°. A lossless transition leads onto a horizontal floor. After 2 m it contacts a light spring fixed to a wall, with stiffness 20 N/m and natural length 1.5 m. The entire horizontal floor has kinetic friction coefficient 0.2. Use g = 10 m/s² and observe until the first stop.
Assumptions and boundaries
- Treat the slider as a point without rotation or size. The incline is smooth; friction acts along the entire horizontal floor, including beneath the spring.
- The join represents a short smooth bend with negligible distance and time, preserving speed and energy. A sharp-corner collision and the transition impulse are not modeled.
- The massless Hookean spring starts uncompressed. End at the first maximum compression or floor stop, before rebound; the last frame is not a claim of permanent equilibrium.
Questions to explore
- What are the speeds at the foot of the ramp and at first spring contact?
- Why is 11 J = kx²/2 not the maximum-compression equation?
- Can it reach the spring if friction becomes 0.8? What if the spring is softer?
Diagram description
Animated physics diagram: ramp_friction_spring; showing time, distance traveled, speed, spring compression, net force along path, normal force, gravitational potential energy, kinetic energy, elastic potential energy, frictional heat.
Unpack one connected process
Use gravitational work to find ramp exit speed. Check that energy remains after the rough run before introducing spring contact. During compression, friction and spring force both act until speed first reaches zero. Carry position, speed and accumulated heat across the same route.
mgh = K + Uspring + Q; Q = μmg × horizontal travel; during compression: ma = −μmg − kx
1. What are the speeds at the foot of the ramp and at first spring contact?
Initial height is 1.5 m and gravitational energy is 15 J. At the foot, v = √30 ≈ 5.477 m/s. Floor friction is 2 N and dissipates 4 J over 2 m. At spring contact, kinetic energy is 11 J and v = √22 ≈ 4.690 m/s. The next stage does not restart from rest.
2. Why is 11 J = kx²/2 not the maximum-compression equation?
Floor friction continues during compression, so 11 = 10x² + 2x. The positive root is x ≈ 0.954 m, below the natural spring length. Elastic energy is about 9.093 J and total frictional heat 5.907 J, still totaling 15 J. Reaching zero speed for the first time does not mean remaining at rest forever.
3. Can it reach the spring if friction becomes 0.8? What if the spring is softer?
At coefficient 0.8, friction is 8 N and stopping distance on the floor is 15/8 = 1.875 m: the slider never reaches the spring. All 15 J becomes heat. Restore friction to 0.2 and change only stiffness to 10 N/m: maximum compression increases to about 1.297 m. Each variation restarts from the same initial state.
Test your prediction in code
Change kinetic friction from 0.2 to 0.8. Can the slider still reach the spring?
scene ramp_friction_spring
body block 1kg point
trajectory trajectory_block body=block stages=(ramp, floor, S) transition=lossless
surface ramp incline 30deg 3m smooth
surface floor horizontal 2m kinetic mu_kinetic=0.2
spring S 20N/m rest_length=1.5m
support wall
attachment attachment_S subject=S target=wall
set gravity 10m/s2
animate block speed=0.36
show block.motion block.velocity block.forces block.energy floor.heat