C/P 05 · Forces, motion and work on an incline
A Cart and a Block on a Ramp
Use Newton’s second law, kinematics and energy to analyze motion on an incline with and without friction.
Passage
Newton’s second law states that the net force on an object equals its mass times its acceleration, F = ma. On an inclined plane that makes an angle θ with the horizontal, it is convenient to separate the weight mg into two components: one parallel to the surface, mg sin θ, which pulls the object down the ramp, and one perpendicular to the surface, mg cos θ, which is balanced by the normal force N. When an object slides, kinetic friction acts up the ramp with a magnitude f = μN, where μ is the coefficient of kinetic friction for the two surfaces.
Energy provides a second way to analyze motion. The work done by a force F acting over a distance d along the direction of motion is W = Fd. The work done by all forces on an object equals its change in kinetic energy, and the work done against friction is converted to thermal energy. For this passage, use g = 10 m/s², sin 37° = 0.6 and cos 37° = 0.8.
Part 1. A student releases a 2 kg cart from rest at the top of a smooth ramp inclined at 37°. The wheels are well lubricated, so friction can be neglected. A motion sensor records the distance the cart has traveled along the ramp at several times. Table 1 shows the results.
Table 1. Distance traveled by the cart on the smooth ramp.
| Time (s) | Distance along the ramp (m) |
|---|---|
| 0.5 | 0.75 |
| 1.0 | 3.0 |
| 1.5 | 6.75 |
| 2.0 | 12.0 |
Part 2. The student replaces the cart with a 2 kg wooden block and covers the ramp with three different surfaces, X, Y and Z. For each surface, the block is placed on the ramp and the acceleration is measured as the block moves down. For surface Z, the block is first given a small push because it does not start to slide on its own; once it is moving, its speed stays constant. Table 2 shows the results.
Table 2. Motion of a 2 kg block on three surfaces on the 37° ramp.
| Surface | Acceleration down the ramp (m/s²) |
|---|---|
| X | 4 |
| Y | 2 |
| Z | 0 (constant speed once moving) |
The student repeats the measurements several times for each surface and finds that the results are reproducible to within a few percent. She also checks that the ramp angle does not change when the block is placed on it, and that the sensor’s reading of the start time matches the release of the block.
On a smooth surface, the only forces on the cart are its weight and the normal force from the ramp, and the normal force does no work because it is perpendicular to the motion. The student therefore expects the acceleration of the cart not to depend on its mass, and she wants to check this by repeating Part 1 with different masses.
Questions
Question 1 of 6
What is the acceleration of the cart on the smooth ramp?
This is the result of dividing the distance at 1.0 s by the time squared and ignoring the factor of ½. Use d = ½at².
This is g, the acceleration of an object in free fall. The cart is on a ramp, so only the component of gravity along the ramp accelerates it.
This is the distance at 2.0 s. It is not an acceleration.
Why the answer is what it is
For constant acceleration from rest, d = ½at². At t = 1.0 s the distance is 3.0 m, so a = 2d/t² = 6 m/s². This agrees with a = g sin 37° = 10 × 0.6 = 6 m/s².
Review: Newtonian Mechanics · Work, Energy, and Power · Equations: Motion · Equations: Energy
Question 2 of 6
What is the speed of the cart at t = 2.0 s?
This is the acceleration. Speed is acceleration times time, v = at.
This is the distance at 2.0 s times 2, and it uses neither v = at nor an average speed.
This would be the speed for a = 1.5 m/s², which does not match the data.
Why the answer is what it is
Starting from rest with a = 6 m/s², v = at = 6 × 2.0 = 12 m/s.
Review: Newtonian Mechanics · Work, Energy, and Power · Equations: Motion · Equations: Energy
Question 3 of 6
The student repeats Part 1 using a 4 kg cart on the same smooth ramp. How will the acceleration compare with that of the 2 kg cart?
The doubled weight is also doubled inertia. The two effects cancel in a = mg sin θ/m.
The component of gravity along the ramp also doubles, and the two effects cancel.
The normal force is perpendicular to the ramp and has no component along the ramp.
Why the answer is what it is
The force along the ramp is mg sin θ and the inertia is m, so a = g sin θ, which is independent of mass.
Review: Newtonian Mechanics · Work, Energy, and Power · Equations: Motion · Equations: Energy
Question 4 of 6
What is the coefficient of kinetic friction for surface X?
This would give an acceleration of 10 × (0.6 − 0.08) = 5.2 m/s², not 4 m/s².
This is the coefficient for surface Y, which gives an acceleration of 2 m/s².
This is the coefficient for surface Z, on which the block moves at constant speed.
Why the answer is what it is
Newton’s second law along the ramp gives ma = mg sin θ − μ mg cos θ, so a = g(sin θ − μ cos θ). For surface X, 4 = 10(0.6 − 0.8μ), so 0.6 − 0.8μ = 0.4 and μ = 0.25.
Review: Newtonian Mechanics · Work, Energy, and Power · Equations: Motion · Equations: Energy
Question 5 of 6
The 2 kg block slides 2.0 m down surface X. How much mechanical energy is converted to thermal energy by friction?
This is the friction force’s magnitude of 4 N, not the work. Work is force times distance.
This is the increase in kinetic energy of the block (net force of 8 N over 2 m). It is not the energy lost to friction.
This is the loss in gravitational potential energy, 12 N × 2 m. Some of it appears as kinetic energy.
Why the answer is what it is
The friction force is f = μmg cos θ = 0.25 × 2 × 10 × 0.8 = 4 N. Over 2.0 m, the work against friction is 4 × 2 = 8 J. The 24 J of gravitational potential energy released equals 16 J of kinetic energy plus 8 J of thermal energy.
Review: Newtonian Mechanics · Work, Energy, and Power · Equations: Motion · Equations: Energy
Question 6 of 6
On surface Z the block moves down the ramp at constant speed. What is the magnitude of the friction force on it?
This is the friction force on surface X plus the net force there. On Z the net force is zero.
This is the normal force, mg cos θ. Friction is not equal to the normal force unless μ = 1.
This is the full weight of the block, mg. Only the component along the ramp must be balanced.
Why the answer is what it is
Constant speed means that the net force along the ramp is zero. Friction therefore balances the component of the weight along the ramp: f = mg sin θ = 2 × 10 × 0.6 = 12 N. This also equals μmg cos θ with μ = 0.75: 0.75 × 16 = 12 N.
Review: Newtonian Mechanics · Work, Energy, and Power · Equations: Motion · Equations: Energy
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