Questions on work, energy and power

Question 1

State the formula for work done by a constant force, and explain why the normal reaction never appears in an energy equation.

Accepted Answer

Work = Fd cos θ, where θ is the angle between the force and the displacement. The normal reaction acts at right angles to the surface and so at right angles to the motion along it, making cos θ = 0 and the work done zero. B1 for the formula, B1 for the normal reaction doing no work.

Question 2

A force of 40 N acts at 25° to the direction of motion while an object moves 12 m. Find the work done.

Accepted Answer

Work = 40 × 12 × cos25° = 480 × 0.9063 = 435 J to three significant figures. M1 for 40 × 12 × cos25°, A1 for 435 J. Ignoring the angle would give 480 J, an overestimate of about 10%.

Question 3

A particle of mass 3 kg has velocity (4i − 2j) m/s. Find its kinetic energy.

Accepted Answer

Kinetic energy = ½m v·v = ½(3)(4² + (−2)²) = ½(3)(20) = 30 J. M1 for v·v as the sum of the squared components, A1 for 30 J. The scalar product of a vector with itself is the square of its magnitude, so this is ½mv² with the speed never written down; energy is a scalar and no direction survives to the answer.

Question 4

A ball of mass 0.5 kg is thrown from a point 5 m above horizontal ground with speed 8 m/s. Air resistance is negligible. Find the speed of the ball when it reaches the ground, taking g = 9.8 m/s².

Accepted Answer

With no resistance, mechanical energy is conserved between the throw and the ground: ½(0.5)v² = ½(0.5)(8²) + 0.5(9.8)(5). M1 for kinetic plus potential energy equated at the two instants. The mass cancels, leaving v² = 64 + 98 = 162; A1. So v = √162 = 12.7 m/s; A1. The angle of projection was not given because it is not needed: gravity does the same 24.5 J of work whichever way the ball is thrown, and only the speed at the ground, not its direction, follows from energy.

Question 5

A particle of mass 60 kg slides from rest 10 m down a slope at 20° to the horizontal, with coefficient of friction 0.3. Find its speed at the bottom, taking g = 9.8 m/s².

Accepted Answer

Work by gravity = 60(9.8)(10)sin20° = 2011 J. Normal reaction = 60(9.8)cos20° = 553 N, so friction is 166 N and the work against it is 1658 J. By the work-energy principle the kinetic energy gained is 2011 − 1658 = 353 J, so ½(60)v² = 353, v² = 11.77 and v = 3.43 m/s. M1 for the work done by gravity, A1 for 2011 J, M1 for the normal reaction and friction, A1 for 1658 J, M1 for the work-energy principle, A1 for 3.43 m/s. Resolve perpendicular to the slope for the normal reaction before touching the friction, since R = mg cos20° rather than mg.

Question 6

A car of mass 900 kg works at 12 kW. At 20 m/s the resistance is 400 N. Find the acceleration then, and the maximum speed if the resistance stays constant.

Accepted Answer

Driving force = 12000/20 = 600 N. Resultant = 600 − 400 = 200 N, so a = 200/900 = 0.222 m/s². At maximum speed the acceleration is zero, so 12000/v = 400 and v = 30 m/s. M1 for P = Fv, A1 for 600 N, M1 for the resultant, A1 for the acceleration, B1 for 30 m/s.

Question 7

Explain why a car's acceleration falls as it speeds up, even though the engine works at constant power.

Accepted Answer

With P fixed, the driving force P/v falls as v rises. The resistance meanwhile grows, so the resultant force shrinks from both ends and the acceleration falls with it. At the speed where the driving force has dropped to equal the resistance, the acceleration reaches zero and the car can go no faster. B1 for the driving force falling, B1 for the resistance rising, B1 for the resultant shrinking to zero.

Question 8

A cyclist and bicycle of total mass 80 kg ride at a steady 6 m/s up a slope of 1 in 20 against a resistance of 25 N. Find the power developed.

Accepted Answer

A slope of 1 in 20 means the component of weight along the slope is 80(9.8)/20 = 39.2 N. At steady speed the driving force balances that plus the resistance: 39.2 + 25 = 64.2 N. Power = Fv = 64.2 × 6 = 385 W to three significant figures. M1 for the component of weight, A1 for 39.2 N, M1 for P = Fv, A1 for 385 W.

Question 9

A particle of mass 3 kg has velocity (2i + 5j) m/s. It moves with constant acceleration (2i − j) m/s² through a displacement of (8i + 8j) m. Find its kinetic energy at the end of that displacement, and the work done on it.

Accepted Answer

No time is given and none is needed: v·v = u·u + 2a·s. u·u = 4 + 25 = 29 and a·s = 16 − 8 = 8, so v·v = 29 + 16 = 45. M1 for v·v = u·u + 2a·s with the scalar products, A1 for 45. The kinetic energy is ½(3)(45) = 67.5 J; A1. It was ½(3)(29) = 43.5 J at the start, so the work done is 67.5 − 43.5 = 24 J; A1. The check is F·s with F = ma = (6i − 3j) N: 48 − 24 = 24 J, the same figure, because v·v = u·u + 2a·s multiplied by ½m is the work-energy principle.

Question 10

A particle of mass 2 kg is projected up a line of greatest slope of a rough plane inclined at 30° to the horizontal, with speed 6 m/s. The coefficient of friction is 0.25. Use the work-energy principle to find the distance the particle travels up the slope before coming to instantaneous rest, taking g = 9.8 m/s².

Accepted Answer

Perpendicular to the slope, R = 2(9.8)cos30° = 16.97 N, so friction is 0.25 × 16.97 = 4.24 N, acting down the slope while the particle moves up. M1 for the normal reaction, A1 for the friction. Over a distance d up the slope, gravity does −2(9.8)(sin30°)d = −9.8d J and friction does −4.24d J, and the kinetic energy falls from ½(2)(6²) = 36 J to zero. M1 for the work-energy equation with both forces opposing the motion. So (9.8 + 4.24)d = 36, that is 14.04d = 36; A1. Hence d = 2.56 m; A1. On a smooth slope the same energy would take the particle 36/9.8 = 3.67 m, so friction has cost just over a metre. Going up, gravity and friction act the same way, which is why their work is added; on the way back down friction reverses and the two would be subtracted.

Question 11

A car of mass 800 kg has an engine that works at a constant 32 kW. The resistance to motion is kv² newtons when the speed is v m/s. The maximum speed of the car on a level road is 40 m/s. Find k, and the acceleration of the car on the level road when its speed is 20 m/s.

Accepted Answer

At the maximum speed the driving force equals the resistance: 32000/40 = k(40²), so 800 = 1600k and k = 0.5. M1 for P/v equal to the resistance at maximum speed, A1 for 0.5. At 20 m/s the driving force is 32000/20 = 1600 N and the resistance is 0.5(20²) = 200 N, so 1600 − 200 = 800a and a = 1.75 m/s². M1 for Newton's second law with the driving force and the resistance both recomputed at 20 m/s, A1 for 1.75 m/s². Halving the speed doubles the driving force and quarters the resistance, which is why the acceleration at 20 m/s is so much larger than the zero at 40. Keeping the resistance at its top-speed value of 800 N gives 1 m/s².

Question 12

A car of mass 1000 kg has an engine working at a constant 20 kW. The resistance to motion is 500 N at all speeds. Find the greatest speed on a level road, the greatest speed up a hill of 1 in 14, and the acceleration up that hill at the moment when the speed is 10 m/s. Take g = 9.8 m/s².

Accepted Answer

On the level the greatest speed comes when the driving force has fallen to the resistance, so 20000/v = 500 and v = 40 m/s.
A hill of 1 in 14 has sin θ = 1/14, so the component of weight down the slope is 1000(9.8)/14 = 700 N. Climbing at a steady speed needs a driving force of 500 + 700 = 1200 N, so 20000/v = 1200 and v = 16.7 m/s.
At 10 m/s the driving force is 20000/10 = 2000 N. Newton's second law along the slope gives 2000 − 500 − 700 = 1000a, so a = 0.8 m/s².
M1 for 20000/v = 500, A1 for 40 m/s, M1 for the component of weight, A1 for 700 N, M1 for the driving force needed, A1 for 16.7 m/s, M1 for Newton's second law, A1 for the acceleration.

Question 13

A particle of mass m moves in a plane with constant acceleration a. Its velocity is u at time 0 and v at time t, and its displacement in that time is s. Starting from v = u + at and s = ut + ½at², prove that v·v = u·u + 2a·s, and show that this relation is the work-energy principle for a constant resultant force F.

Accepted Answer

Take the scalar product of v = u + at with itself: v·v = (u + at)·(u + at). M1 for forming v·v as a scalar product. The scalar product distributes over addition and u·a = a·u, so v·v = u·u + 2t(u·a) + t²(a·a). A1.
Now 2a·s = 2a·(ut + ½at²) = 2t(a·u) + t²(a·a). M1 for expanding a·s with the same rules. This is exactly the part of v·v after u·u, so v·v = u·u + 2a·s. A1.
Multiply through by ½m: ½m v·v − ½m u·u = m a·s = (ma)·s = F·s, using F = ma for a constant resultant force. M1 for multiplying by ½m and replacing ma by F. The left side is the change in kinetic energy, ½m v·v being ½mv² with v² read as v·v, and the right side is the work done by F over the displacement s, since F·s = |F||s|cos θ is force times distance along the force. So the relation says work done equals change in kinetic energy, which is the work-energy principle. A1.
Of the five constant-acceleration formulae this is the only one that needs a scalar product to cross into vectors, because it is the only one that squares a vector; the others are linear and carry over term by term. Writing v² = u² + 2as with vectors in place of the letters is not meaningful, since a vector cannot be squared or multiplied by another without saying which product is meant.

Question 14

A car of mass 1200 kg has an engine working at a constant 30 kW. The resistance to motion is proportional to the speed, and the car's maximum speed on a level road is 50 m/s. Find the maximum speed of the car up a hill inclined at angle α to the horizontal, where sin α = 1/20, taking g = 9.8 m/s².

Accepted Answer

Write the resistance as kv. On the level at maximum speed the driving force equals the resistance: 30000/50 = 50k, so 600 = 50k and k = 12. M1 for P/v equal to the resistance at 50 m/s, A1 for k = 12.
Up the hill the component of weight down the slope is 1200(9.8)/20 = 588 N. At the maximum speed V the driving force balances resistance and weight component: 30000/V = 12V + 588. M1 for the balance up the hill with both terms. Multiplying by V: 12V² + 588V − 30000 = 0, that is V² + 49V − 2500 = 0. A1 for the quadratic.
V = [−49 + √(49² + 10000)]/2 = (−49 + 111.36)/2, taking the positive root since a speed cannot be negative; M1 for solving with the negative root rejected. So V = 31.2 m/s; A1.
The resistance at 31.2 m/s is 374 N, well below the 600 N on the level at 50 m/s, so the hill has not added 588 N on top of the old resistance: as the speed falls the resistance falls with it, and the driving force 30000/V rises. That is why the answer needs a quadratic and is not 30000/(600 + 588) = 25.3 m/s, which treats the resistance as fixed at its level-road value.

Question 15

A vehicle of mass m moves on a level road with its engine working at a constant power P against a constant resistance R. Show that when its speed is v its acceleration is (P − Rv)/(mv). Deduce the set of speeds at which the vehicle is accelerating, and explain why, starting from rest, it approaches its maximum speed but never reaches it.

Accepted Answer

At speed v the driving force is F = P/v, from P = Fv. M1 for the driving force. Newton's second law along the road gives P/v − R = ma, so a = (P/v − R)/m = (P − Rv)/(mv). A1.
Since m and v are positive, a > 0 exactly when P − Rv > 0, that is when v < P/R. M1 for the sign of the numerator deciding the sign of a, A1 for v < P/R.
Starting from rest the speed rises, and while v < P/R the acceleration stays positive, so v keeps rising; but as v approaches P/R the numerator P − Rv approaches zero and the acceleration shrinks towards zero. The speed can never pass P/R, because the acceleration would be negative beyond it, and it cannot reach P/R either, since at every speed below it the acceleration is positive but becoming smaller, so the remaining gap is closed more and more slowly. P/R is a limit the speed tends to, not a value it attains. B1 for the argument that a tends to zero as v tends to P/R, so the maximum speed is approached and not reached.
The model also predicts an infinite acceleration at v = 0, since P/v is unbounded there; a real engine cannot deliver its full power from rest, so the constant-power model is used once the vehicle is already moving.