ANSWERS to the kinetic energy store problems

Q1 A swimmer of mass 70 kg is moving at a speed of 1.4 m/s.

Calculate the kinetic energy of the swimmer in J (to 2 sf).

ANSWERS

 

Q2 A bullet with a mass of 10.0 g is travelling with an initial speed of 400 m/s.

(a) What is the bullet's Initial kinetic energy store?

(b) If the bullet embeds itself in a wooden plank, by how much will the planks thermal energy store be increased?

ANSWERS

 

Q3 A car of mass 1200 kg is travelling at a steady speed of 30.0 m/s (~67 mph, 108 km/hour).

(a) What is the kinetic energy store of the car?

(b) If on slowing down the kinetic energy of the car is halved, what speed is it then doing?

ANSWERS

 

Q4 The chemical potential energy store of a 100 g rocket, on firing, gives it an initial kinetic energy store of 200 J.

Calculate the initial vertical velocity of the rocket.

ANSWERS

 

Q5 A projectile is required to impart a kinetic energy blow of 5000 J at a speed of 500 m/s.

ANSWERS

 

Q6. A pole vaulter has a weight of 800 N and vaults to a height of 4.5 m.

(See gravitational potential energy calculation page)

(a) How much work does the pole vaulter do?

(c) At the maximum height reached, what gravitational potential energy does the pole vaulter possess?

(b) What kinetic energy did the pole vaulter impart to its body on 'take-off' (and what assumption are you making?)

(c) What is the kinetic energy of the pole vaulter immediately before impacting on the ground?

(d) What was the initial speed of take-off by the pole vaulter? (take gravity strength as 9.8 N/kg).

ANSWERS

 

Q7 A diver of mass 60 kg is perched on a 7.5 m high diving board. (First see GPE calculations page)

In your GPE calculations take gravitational field strength acceleration to be 9.8 m/s².

The 10 m diving board is the highest in Olympic sport, 7.5 m board is the next highest.

(a) Calculate the gravitational potential energy (GPE) of the diver with respect to the surface of the water in the pool.

(b) Just before the diver hits the water surface, what is the diver's maximum kinetic energy? Explain your answer.

(c) From your answers to (a) and (b) calculate the diver's speed on impacting on the surface of the water.

ANSWERS

 

Q8 A small bee of mass 0.06 g leaps up 10 cm above a flower. (g = 10 N/kg)

(a) Calculate the gain in gravitational potential energy.

(b) Assuming there is no further acceleration, and ignoring air resistance, what was the initial take off speed of the bee?

(c) If the same bee leaps from another flower and gains 2.0 x 10^-4 J, how high did the bee rise?

ANSWERS

 

Q9 In falling, an initially stationary 2.00 kg weight, loses 500 J of energy before hitting the floor.

(a) At what height was the weight dropped? (assume g = 10 N/kg)

(b) Neglecting air resistance, at what speed did the 2 kg weight hit the floor?

ANSWERS

 

Q10 An electric motor in a toy train does 0.2 J of work to accelerate it to a speed of 25 cm/s.

(a) If the toy train has a mass of 3500 g, what is the efficiency of the motor?

(b) Why is the efficiency much less than 100%?

ANSWERS

Keywords, phrases and learning objectives on kinetic energy stores

Be able to explain with examples what a kinetic energy store is.

Be able to explain how a kinetic energy store is converted into another energy store e.g. thermal energy from friction, electrical energy from a generator.

Know how to do calculations and solve problems using the formula for kinetic energy.

Kinetic energy store problems

Q1 A swimmer of mass 70 kg is moving at a speed of 1.4 m/s.

Calculate the kinetic energy of the swimmer in J (to 2 sf).

E_ke = ¹/₂ m v² = 0.5 x 70 x 1.4² = 68.6 = 69 J (2 sf)

 

Q2 A bullet with a mass of 10.0 g is travelling with an initial speed of 400 m/s.

(a) What is the bullet's Initial kinetic energy store?

E_ke = ¹/₂ m v² , 10 g ≡ 10/1000 = 0.01 kg

E_ke = 0.5 x 0.01 x 400²

E_ke = 800 J of kinetic energy

(b) If the bullet embeds itself in a wooden plank, by how much will the planks thermal energy store be increased?

All of the 800 J of kinetic energy will end up as heat. There will be some loss due to friction (air resistance) and sound energy, but most of the 800 J will end up as heat energy in the plank.

 

Q3 A car of mass 1200 kg is travelling at a steady speed of 30.0 m/s (~67 mph, 108 km/hour).

(a) What is the kinetic energy store of the car?

E_ke = ¹/₂ m v² , 10 g ≡ 10/1000 = 0.01 kg

E_ke = 0.5 x 1200 x 30²

E_ke = 540,000 J of kinetic energy (5.40 x 10⁵J, 0.540 MJ, 3 s.f.)

(b) If on slowing down the kinetic energy of the car is halved, what speed is it then doing?

E_ke = ¹/₂ m v² , rearranging:   v² = 2E_ke / m,  so v = √(2E_ke / m),  KE = 540,000/2 = 270,000 J

v = √(2 x E_ke / m) = √(2 x 270,000 / 1200) = 21.2 m/s (3 s.f.)

 

Q4 The chemical potential energy store of a 100 g rocket, on firing, gives it an initial kinetic energy store of 200 J.

Calculate the initial vertical velocity of the rocket.

E_ke = ¹/₂ m v² , rearranging:   v² = 2E_ke / m,  so v = √(2E_ke / m)

m = 100 g ≡ 100 / 1000 = 0.1 kg

v = √(2 x E_ke / m) = √(2 x 200 / 0.10) = 63.2 m/s (3 s.f.)

 

Q5 A projectile is required to impart a kinetic energy blow of 5000 J at a speed of 500 m/s.

What mass of projectile is required?

E_ke = ¹/₂ m v² , rearranging:   m = 2 x E_ke / v²

m = 2 x E_ke / v² = 2 x 5000 / 500² = 0.040 kg (40.0 g, 3 s.f.)

 

Q6. A pole vaulter has a weight of 800 N and vaults to a height of 4.5 m. (See gravitational potential energy calculation page)

(a) How much work does the pole vaulter do?

work (J) = force (N) x distance of action (m)

work done = 800 x 4.5 = 3600 J

(c) At the maximum height reached, what gravitational potential energy does the pole vaulter possess?

The GPE equals the work done in raising the pole vaulter to a height of 4.5 m, that is 3600 J

(b) What kinetic energy did the pole vaulter impart to its body on 'take-off' (and what assumption are you making?)

If you ignore air resistance i.e. no energy lost - wasted, the initial KE = maximum GPE = 3600 J

(c) What is the kinetic energy of the pole vaulter immediately before impacting on the ground?

maximum KE = maximum GPE gained = 3600 J (neglecting air resistance energy losses)

(d) What was the initial speed of take-off by the pole vaulter? (take gravity strength as 9.8 N/kg).

KE = ¹/₂mv², rearranging gives v = √(2KE/m)

initial KE = 3600 J, mass = 800 / 9.8 = 81.63 kg

v = √(2KE/m) = √(2 x 3600/81.63) = 9.4 m/s (2sf)

 

Q7 A diver of mass 60 kg is perched on a 7.5 m high diving board. (First see GPE calculations page)

In your GPE calculations take gravitational field strength acceleration to be 9.8 m/s².

The 10 m diving board is the highest in Olympic sport, 7.5 m board is the next highest.

(a) Calculate the gravitational potential energy (GPE) of the diver with respect to the surface of the water in the pool.

GPE = m g h = 60 x 9.8 x 7.5 = 4410 J

(b) Just before the diver hits the water surface, what is the diver's maximum kinetic energy? Explain your answer.

From the law of conservation of energy, the GPE is converted into kinetic energy (neglecting air resistance).

Therefore the diver's maximum kinetic energy is also 4410 J

(c) From your answers to (a) and (b) calculate the diver's speed on impacting on the surface of the water.

KE = ½ mv², rearranging gives: v = √(2KE / m)

v = √(2 x 4410 / 60) = √147 = 12 m/s (2 s.f.)

 

Q8 A small bee of mass 0.06 g leaps up 10 cm above a flower. (g = 10 N/kg)

(a) Calculate the gain in gravitational potential energy.

∆E = mgh = (0.0.06/1000) x 10 x (10/100)

= 6 x 10^-5 x 10 x 10^-1 =  6.0 x 10^-5 J  (2 sf)

(b) Assuming there is no further acceleration, and ignoring air resistance, what was the initial take off speed of the bee?

As the bee ascends, the kinetic energy store of the bee is converted into the GPE store of the bee,

Therefore final maximum ∆GPE = initial maximum ∆KE

So, ∆GPE = ∆KE = 6.0 x 10^-5 J

KE = ¹/₂mv², rearranging gives v = √(2KE/m)

v = √(2KE/m) = √(2 x 10^-5 / 6 x 10^-5) = √0.3333 = 0.58 m/s  (2 sf)

(c) If the same bee leaps from another flower and gains 2.0 x 10^-4 J, how high did the bee rise?

∆GPE = mgh, rearranging gives h = ∆GPE / mg

h = 2.0 x 10^-4 / (6.0 x 10^-5 x 10) = 0.33 m  (33 cm, 2 sf)

 

Q9 In falling, an initially stationary 2.00 kg weight, loses 500 J of energy before hitting the floor.

(a) At what height was the weight dropped? (assume g = 10 N/kg)

The loss of energy is equal to the loss of GPE as it is converted to KE.

∆GPE = mgh, rearranging gives:

h = ∆GPE / (g x m) = 500 / (10 x 2) = 500 / 20 = 25 m

(b) Neglecting air resistance, at what speed did the 2 kg weight hit the floor?

Assuming maximum ∆GPE = ∆KE on impact

KE = ¹/₂mv², rearranging gives v = √(2KE/m)

v = √(2KE/m) = √(2 x 500 /2) = √500 = 22.4 m/s  (3 sf)

Q10 An electric motor in a toy train does 0.2 J of work to accelerate it to a speed of 25 cm/s.

(a) If the toy train has a mass of 3500 g, what is the efficiency of the motor?

First calculate the kinetic energy store of the train.

mass = 3500/1000 = 3.50 kg, speed = (25/100) = 0.25 m/s

KE = ¹/₂mv²  = 0.5 x 3.5 x 0.25² = 0.1094 J

Efficiency = useful energy out / total energy input = 0.1094 / 0.20 = 0.55 (2 s.f.)

Efficiency = 0.55 x 100 = 55% (2 s.f.)

(b) Why is the efficiency much less than 100%?

Apart from a little sound, most energy is lost (dissipated) from the friction of the moving parts of the electric motor and friction between the wheels and the rails - all energy lost ends up increasing the thermal energy store of the surroundings.

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