Cart on a Ramp: Where Did the Energy Go?
At a Fukui prefectural science fair, a team investigates energy conversions using a 2.0 kg steel cart. The cart is released from rest at the top of a straight ramp 3.0 m long whose upper end is 1.2 m above the floor. A light gate at the bottom records 3.6 m/s. The cart then runs onto a level floor and comes to rest 3.3 m from the foot of the ramp. The team is told to use g = 9.8 m/s^2 and to ignore air resistance and the small energy stored in the cart's wheels. Their notebook contains three entries: 'gravitational potential energy lost = 23.5 J', 'speed at the bottom should be about 4.9 m/s if nothing is wasted', and a partner's remark that the cart 'uses up force as it rolls'. The teacher asks them to account for the missing energy quantitatively and to decide whether one friction model can explain both the ramp and the floor sections.
1. Calculate the energy converted to thermal energy on the ramp and the average friction force acting along the ramp. State the direction of that force.
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Energy accounting. The cart starts from rest, so initial kinetic energy is zero. Potential energy lost = mgh = 2.0 x 9.8 x 1.2 = 23.5 J. Kinetic energy at the bottom = 1/2 mv^2 = 0.5 x 2.0 x 3.6^2 = 13.0 J. The difference, 23.5 - 13.0 = 10.5 J (about 11 J), is no longer mechanical energy; it has become thermal energy in the ramp surface, the cart and the air.
Force. Average friction force = energy / distance = 10.5 / 3.0 = 3.5 N, directed up the slope, opposite to the motion. This is an average: real friction may vary along the ramp, and a little energy also went into rotation of the wheels, so 3.5 N is an upper estimate for sliding friction alone.
2. On the floor the cart travels 3.3 m before stopping. Find the friction force there and judge whether a single coefficient of friction is consistent with both sections.
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Floor section. The cart arrives with 13.0 J of kinetic energy and loses all of it over 3.3 m, so the friction force is 13.0 / 3.3 = 3.9 N.
Comparing coefficients. On the ramp the normal force is mg cos(theta). The slope angle satisfies cos(theta) = sqrt(3.0^2 - 1.2^2) / 3.0 = 2.75 / 3.0 = 0.917, so N = 2.0 x 9.8 x 0.917 = 18.0 N and mu = 3.5 / 18.0 = 0.19. On the floor N = 2.0 x 9.8 = 19.6 N and mu = 3.9 / 19.6 = 0.20.
Judgement. The two coefficients agree within reading uncertainty, so one friction model does explain both sections. The friction force on the ramp is smaller only because the normal force there is smaller, not because the surfaces differ. A claim that 'the ramp has less friction' misreads the data: the coefficient, not the force, describes the surfaces.
3. The partner predicts that a ramp twice as high would send the cart off at twice the speed, and says the cart 'uses up force'. Evaluate both statements.
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Speed scaling. Without friction, v = sqrt(2gh), so doubling h multiplies v by sqrt(2) = 1.41, not 2. From 1.2 m: sqrt(2 x 9.8 x 1.2) = 4.85 m/s; from 2.4 m: 6.86 m/s.
With friction the answer depends on how the ramp is changed. If the slope angle is kept and the height doubled, the ramp is twice as long, friction loss doubles along with the energy input, kinetic energy still doubles and the speed gain is still about 1.41x. If instead only the height is doubled while the 3.0 m length is kept, the ramp is steeper, the normal force is smaller, the friction loss is slightly less, and the gain is slightly above 1.41x - but never 2x.
'Uses up force'. This has no meaning: force is an interaction, not a fuel. What the cart uses up is kinetic and gravitational potential energy, converting it into thermal energy. The force of friction is the mechanism of that conversion, not something consumed.