Block on an Incline Compressing a Spring with Friction
A 2.0 kg block is released from rest at the top of a straight incline that makes 30° with the horizontal. The block slides 3.0 m along the slope before reaching a smooth, short transition to a horizontal section. The coefficient of kinetic friction between the block and both surfaces is 0.25. On the horizontal section, the block meets a spring of spring constant 200 N/m and compresses it before stopping momentarily. The spring is aligned with the motion, and the block does not leave the surface. A student first calculates the speed at the bottom of the incline by assuming no friction, then is asked to redo the calculation with friction and to decide whether friction during the spring compression can be neglected. Use g = 9.8 m/s². The transition is short enough that its length can be ignored, but the student is warned that 'ignoring the transition' is not the same as 'ignoring friction on the horizontal part'.
1. Calculate the speed of the block at the bottom of the incline. State whether you used forces or work-energy and why.
Read the worked answer
Work-energy is more direct because the block moves along a known distance and the forces are constant. The normal force on the incline is
[N=mg\cos30^\circ=(2.0)(9.8)(0.866)=16.97\ \text{N}.]
So kinetic friction is
[f_k=\mu_kN=(0.25)(16.97)=4.24\ \text{N}.]
The vertical drop is (h=3.0\sin30^\circ=1.5\ \text{m}), so gravity does
[W_g=mgh=(2.0)(9.8)(1.5)=29.4\ \text{J}.]
Friction does
[W_f=-f_k(3.0)=-(4.24)(3.0)=-12.7\ \text{J}.]
Thus the kinetic energy at the bottom is
[K=29.4-12.7=16.7\ \text{J},]
and
[\frac12(2.0)v^2=16.7\Rightarrow v\approx4.1\ \text{m/s}.]
The frictionless answer would be (v=\sqrt{2gh}=5.4\ \text{m/s}), so friction removes a substantial fraction of the energy.
2. Calculate the maximum compression x of the spring, including friction on the horizontal section. Explain why ignoring friction during compression gives a different value.
Read the worked answer
At the bottom of the incline the block has (16.7\ \text{J}) of kinetic energy. On the horizontal section the normal force is (mg=19.6\ \text{N}), so friction is
[f_k=(0.25)(19.6)=4.90\ \text{N}.]
While the spring compresses by (x), friction also does negative work (-4.90x). Energy conservation gives
[16.7=\frac12(200)x^2+4.90x.]
So
[100x^2+4.90x-16.7=0.]
The positive root is
[x=\frac{-4.90+\sqrt{4.90^2+4(100)(16.7)}}{200} \approx\frac{-4.90+81.8}{200}\approx0.385\ \text{m}.]
If friction during compression were ignored, the result would be (x=\sqrt{16.7/100}\approx0.409\ \text{m}). The difference is about 0.024 m, or roughly 6%. Friction removes energy during the compression itself, so less energy is available to store in the spring.
3. After rebounding, the block is observed to travel 1.2 m up the incline. Without a full new calculation, judge whether this is consistent with μk = 0.25, and identify one additional loss mechanism that could matter.
Read the worked answer
The spring stores (\frac12(200)(0.385)^2\approx14.8\ \text{J}). On the return to the equilibrium position, friction removes another (4.90(0.385)\approx1.9\ \text{J}), leaving at most about (12.9\ \text{J}) of kinetic energy. Even if friction on the return were ignored, the available energy would be only (14.8\ \text{J}). Going up the incline costs (mg\sin30^\circ=9.8\ \text{N}) per metre against gravity plus (4.24\ \text{N}) per metre against friction, or about (14.0\ \text{J/m}). Thus the maximum possible distance is at most (14.8/14.0\approx1.05\ \text{m}), and with friction it is about (12.9/14.0\approx0.92\ \text{m}). A 1.2 m rise would require more energy than the spring can supply, so the observation is not consistent with (\mu_k=0.25). Possible additional losses include sound, deformation of the block or surface, and internal friction or hysteresis in the spring; these would make the distance even shorter, not longer.