Energy, Work & Momentum

11 methods · 25 questions

★ Signature method unlocked: the technique this chapter rewards most often, with 3 worked questions.

The full chapter (11 methods · 25 worked questions) comes with Platform Access. See pricing

When a question asks for energy lost, energy wasted, or work done against a resistive force, do not model the forces: draw up one balance sheet over the whole journey, everything that went in on one side, the useful outcome on the other, and the difference is the answer.

Trigger: the wording lost to the surroundings, wasted, work done against drag, or work done against resistive forces.

Instances:
(i) a fall with drag: $W_{drag} = (E_{k, i} + m g h) - E_{k, f}$ , in one line, with no force and no SUVAT;
(ii) a horizontal push: $W_{resist} = F d - Δ E_{k}$ ;
(iii) a machine lifting a load: energy lost $=$ energy supplied $- m g h$ , and an efficiency quoted for one stage of the machine is a distractor, since the balance already contains every loss;
(iv) lifted slowly, or at constant speed, means there is no kinetic energy term at all and the useful output is exactly $m g h$ .

Linked questions (3)

NSAA 2017 Section 1 Part B (ESAT Physics) · Q23

An electric motor is used to pull a broken-down car slowly from the road up a ramp on to the back of a breakdown truck.

The car has a mass of

1200 kg

and is lifted through a vertical height of

1.0 m

.

The total input energy to the motor is

28 kJ

and it is

75%

efficient.

In the process of lifting the car, energy is lost to the surroundings from the motor and from other causes.

What is the total energy lost to the surroundings?

(gravitational field strength

= 10 N kg^{- 1}

)

A $7.0 kJ$
B $9.0 kJ$
C $12 kJ$
D $16 kJ$
E $21 kJ$
F $33 kJ$

Worked solution

Answer:D

The only useful outcome of the whole process (motor plus ramp) is the gain in gravitational potential energy of the car, since it is lifted 'slowly' (negligible kinetic energy gained). This useful energy is

E_{p} = m g h = 1200 \times 10 \times 1.0 = 12000 J = 12 kJ .

By conservation of energy, everything else supplied to the motor that does not end up as this useful PE must be lost to the surroundings, whether inside the motor (the stated

75%

efficiency) or elsewhere in the lifting process (friction on the ramp, etc.):

E_{lost} = E_{input} - E_{p} = 28 - 12 = 16 kJ .

(Note that the motor's own

75%

efficiency figure,

0.75 \times 28 = 21 kJ

, is not itself the answer: it is only the energy leaving the motor as useful mechanical work, some of which is then further lost to 'other causes' before the car's PE is gained. The total-energy-balance approach above sidesteps needing that intermediate step.)

The answer is D.

NSAA 2020 Section 1 Part B (ESAT Physics) · Q38

A parachutist of mass

80.0 kg

drops from a plane travelling at

40.0 m s^{- 1}

2000 m

above the Earth's surface.

The parachutist hits the ground at a speed of

5.00 m s^{- 1}

.

How much work is done by the parachutist against drag forces during the fall?

(Take the Earth's gravitational field strength to be

10.0 N kg^{- 1}

A $1535000 J$
B $1624000 J$
C $1649000 J$
D $1663000 J$
E $1726000 J$

Worked solution

Answer:D

By conservation of energy, the initial kinetic and gravitational potential energy convert into final kinetic energy plus the work done against drag:

E_{k, i} + E_{p, i} = E_{k, f} + E_{p, f} + W_{drag} .

Taking the ground as the reference level (

E_{p, f} = 0

E_{p, i} = m g h = (80.0 kg) (10.0 N kg^{- 1}) (2000 m) = 1600000 J .

E_{k, i} = \frac{1}{2} m v_{i}^{2} = \frac{1}{2} (80.0 kg) (40.0 m s^{- 1})^{2} = 64000 J .

E_{k, f} = \frac{1}{2} m v_{f}^{2} = \frac{1}{2} (80.0 kg) (5.00 m s^{- 1})^{2} = 1000 J .

W_{drag} = (E_{k, i} + E_{p, i}) - E_{k, f} = (64000 + 1600000) - 1000 = 1663000 J .

The answer is D.

NSAA 2021 Section 1 Part B (ESAT Physics) · Q36

A skydiver of mass

80 kg

is accelerating vertically downwards through the air. At one instant in time the skydiver has a speed of

5.0 m s^{- 1}

. After travelling a further distance of

20 m

downwards the skydiver's speed has increased to

10 m s^{- 1}

.

What is the average force of air resistance acting on the skydiver over the

20 m

?

(gravitational field strength

= 10 N kg^{- 1}

)

A $600 N$
B $650 N$
C $750 N$
D $790 N$
E $950 N$

Worked solution

Answer:B

Using the work-energy theorem over the

20 m

fall, the change in kinetic energy equals the net work done:

Δ E_{k} = \frac{1}{2} (80 kg) [(10 m s^{- 1})^{2} - (5.0 m s^{- 1})^{2}] = \frac{1}{2} (80) (100 - 25) = 3000 J .

The work done by gravity over the fall is

W_{gravity} = m g h = (80 kg) (10 N kg^{- 1}) (20 m) = 16000 J .

Air resistance acts upward (opposing the downward motion), doing negative work

- F_{avg} \times 20 m

. The net work equals the kinetic energy change:

W_{gravity} - F_{avg} (20 m) = Δ E_{k}

16000 J - F_{avg} (20 m) = 3000 J

F_{avg} = \frac{16 000 - 3000}{20} N = \frac{13 000}{20} N = 650 N .

The answer is B.