A locomotive of mass m starts moving so that its velocity varies according to the law v = a√s where a is a constant, and s is the distance covered. Find the total work performed by all the forces which are acting on the locomotive during the first t seconds after the beginning of motion.
The total work of all the forces in the first t seconds is A = m a⁴ t²/8. With the numbers on screen, A = ( J).
Problem statement
Problem 1.80. A locomotive of mass m starts moving so that its velocity varies according to the law v = a√s where a is a constant, and s is the distance covered. Find the total work performed by all the forces which are acting on the locomotive during the first t seconds after the beginning of motion.
We assume the locomotive starts from rest at s = 0 and moves in a straight line. It is a slider for the time t.
We assume m = 50 tonnes (50 000 kg) and a = 1 m½/s, because the problem gives no numbers. The unit of a is m½/s, since v is in m/s and √s is in m½. Both are sliders.
We assume t = 10 s as the starting time. It is a slider.
Symbols: w is the acceleration, F the net force, s the distance covered, v the speed, A the work.
The big idea in plain words
The rule v = a√s looks odd, but it hides a simple fact. The speed grows as the square root of the distance. When you work out how fast the speed changes with time, the root cancels, and the acceleration comes out constant. A constant acceleration means a constant force. Then the work is just force times distance.
The solution in each frame
Only the ground frame is needed. The track is at rest in it, and the problem asks for the work of the forces in that frame. A frame fixed to the locomotive would accelerate and would show no motion at all.
The solution step by step
Step 1, the acceleration. The locomotive moves in one direction, so w = dv/dt. Differentiate v = a√s: dv/dt = (dv/ds)(ds/dt) = (a/(2√s))(a√s) = a²/2. So w = a²/2 = m/s², and it does not change.
Step 2, the force. By Newton's second law the net force is F = mw = ma²/2 = kN.
Step 3, the distance. The acceleration is constant and the start is from rest, so s = ½wt² = a²t²/4 = m.
Step 4, the work. The force is constant and along the motion, so A = Fs = (ma²/2)(a²t²/4) = ma⁴t²/8 = .
Step 5, the cases. The problem has one situation, so no cases are needed. Two results follow from the formula. Doubling the time makes the work four times as large. Doubling a makes the work sixteen times as large.
A separate calculation does not use the formula. It solves ds/dt = a√s on the computer in 200000 small steps, with a tiny start of 10−12 m. It gets s = m, against m from the formula. It then finds the work in two other ways. Half of m times v squared gives J, and adding up m(dv/dt)Δs over the steps gives J, against J from the formula. The largest relative difference is .
Five methods
1. Force method
Find the acceleration, then F = mw, then A = Fs. This is steps 1 to 4.
2. Energy method
The work of all the forces equals the change of kinetic energy. At the end, v = a√s = a²t/2, so A = ½mv² = ½m(a²t/2)² = ma⁴t²/8. The same result, without finding the acceleration.
3. Graph method
The third graph shows v against t. It is a straight line, so the acceleration is constant. The area under the force against distance is the work, and that force is constant, so the area is a rectangle F × s. The second graph shows A against t on log scales. It is a straight line with slope 2, which means A is proportional to t².
4. Calculus method
A = ∫F ds = ∫m(dv/dt)ds = ∫mv dv = ½mv², which gives the energy result again.
5. Check method
Units: a⁴ is in m²/s&sup4; and t² in s², so m a⁴ t² has the units kg m²/s², which is joules. Limits: if t is zero the work is zero. The work grows as t², because both the force-distance and the speed grow with time.
Fastest: the energy method. Most reliable: the force method with the numerical check.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Mass m | Yes | The force and the work. Not the acceleration or the distance |
| Constant a | Yes | The acceleration, distance, speed, force, work and all curves |
| Time t | Yes | The distance, speed, work and the picture |
| Given-answer mode | Sets t | The page finds t for a chosen work and warns if t would be below 1 s or above 20 s |
| Play and time bar | No | Play runs the motion, holds 1.2 s at the end, then repeats. The bar sets the time |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Mass | m | t ( kg) |
| Constant in v = a√s | a | m½/s |
| Time | t | s |
| Acceleration | w = a²/2 | m/s² |
| Force | F = mw | kN |
| Distance | s | m |
| Speed at time t | v | m/s |
| Work | A |
Exam traps
- Thinking the force changes because the speed depends on the distance. The acceleration is constant, a²/2.
- Using A = Fs with an F that was found from v alone. Differentiate in time first, not in distance.
- Using a wrong formula for the distance. With constant w from rest, s = ½wt² = a²t²/4, not at.
- Forgetting to square a twice. The work has a to the fourth power.
- Giving the work per second. The question asks for the total work in the first t seconds.