EduNinja

Learn this topic

Study A.3 Work, energy and power conceptsOpen the corresponding Concept lesson for explanations, examples and exam connections.

[Maximum number: 5]

This question is in two parts. Part 1 is about energy resources. Part 2 is about thermal physics.
Part 1 Energy resources
Electricity can be generated using nuclear fission, by burning fossil fuels or using pump storage hydroelectric schemes.

(a)

A hydroelectric scheme has an efficiency of 92 %. Water stored in the dam falls through an average height of 57 m . Determine the rate of flow of water, in $\mathrm{kg} \mathrm{s}^{-1}$ , required to generate an electrical output power of 4.5 MW .

[ 3 ]

use of $\frac{m g h}{t}$ ;
$\frac{m}{t}=\frac{4.5 \times 10^{6}}{0.92 \times 9.81 \times 57}$ ;
$8.7 \times 10^{3} \mathrm{~kg} \mathrm{~s}^{-1}$ ;

(b)

A mass of 0.22 kg of lead spheres is placed in a well-insulated tube. The tube is turned upside down several times so that the spheres fall through an average height of 0.45 m each time the tube is turned. The temperature of the spheres is found to increase by $8^{\circ} \mathrm{C}$ .

[ 2 ]

(i)

Discuss the changes to the energy of the lead spheres.

[ 2 ]

[Maximum number: 11]

This question is in two parts. Part 1 is about the motion of a ship. Part 2 is about melting ice.

(a)

Outline the meaning of work.

[ 2 ]

work done = force × distance moved;
(distance moved) in direction of force;
or
energy transferred;
from one location to another;
or
work done $=F s \cos \theta$ ;
with each symbol defined;

(b)

Some cargo ships use kites working together with the ship's engines to move the vessel.

The tension in the cable that connects the kite to the ship is 250 kN . The kite is pulling the ship at an angle of $39^{\circ}$ to the horizontal. The ship travels at a steady speed of $8.5 \mathrm{~m} \mathrm{~s}^{-1}$ when the ship's engines operate with a power output of 2.7 MW .

[ 6 ]

(i)

Calculate the work done on the ship by the kite when the ship travels a distance of 1.0 km .

[ 2 ]

horizontal force $=250000 \times \cos 39^{\circ}\left(=1.94 \times 10^{5} \mathrm{~N}\right)$ ;
work done $=1.9 \times 10^{8} \mathrm{~J}$ ;

(ii)

Show that, when the ship is travelling at a speed of $8.5 \mathrm{~m} \mathrm{~s}^{-1}$ , the kite provides about 40 % of the total power required by the ship.

[ 4 ]

power provided by kite $=\left(1.94 \times 10^{5} \times 8.5=\right) 1.7 \times 10^{6} \mathrm{~W}$ ;
total power $=(2.7+1.7) \times 10^{6} \mathrm{~W} \quad\left(=4.4 \times 10^{6} \mathrm{~W}\right)$ ;
fraction provided by kite $=\frac{1.7}{2.7+1.7}$ ;
38 % or 0.38 ;
(must see answer to 2+ sig figs as answer is given)

Marking guidance:

Allow answers in the range of 37 to 39 % due to early rounding.
or
Award [3 max] for a reverse argument such as:
if 2.7 MW is 60 %;
then kite power is $\frac{2}{3} \times 2.7 \mathrm{MW}=1.8 \mathrm{MW}$ ;
shows that kite power is actually 1.7 MW ;
(QED)

(c)

The kite is taken down and no longer produces a force on the ship. The resistive force F that opposes the motion of the ship is related to the speed v of the ship by

F=k v^{2}

where k is a constant.
Show that, if the power output of the engines remains at 2.7 MW , the speed of the ship will decrease to about $7 \mathrm{~ms}^{-1}$ . Assume that k is independent of whether the kite is in use or not.

[ 3 ]

$\quad P=\left(k v^{2}\right) \times v=k v^{3}$ ;
$\frac{v_{1}}{v_{2}}=\left(\sqrt[3]{\left(\frac{P_{1}}{P_{2}}\right)}=\right) \sqrt[3]{\left(\frac{2.7}{4.4}\right)} ;$
final speed of ship $=7.2 \mathrm{~ms}^{-1}$ ; (at least 2 sig figs required).

Approximate answer given, marks are for working only.

[Maximum number: 11]

(a)

The conveyor belt moves with a constant horizontal speed of $1.5 \mathrm{~ms}^{-1}$ . As the gravel lands on the belt, it has no horizontal speed.

[ 3 ]

(i)

Calculate the rate of change of the kinetic energy of the gravel due to its change in horizontal speed.

[ 1 ]

$14.6 \mathrm{~J} \mathrm{~s}^{-1}$ ;

(ii)

Determine the power required to move the conveyor belt at constant speed.

[ 2 ]

horizontal momentum gain per second $=13 \times 1.5\left(=19.5 \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}\right)$ ;
power required $=29.3 \mathrm{~W}$ ;

(b)

Oscillating water column (OWC) energy converters placed in the ocean are suggested for non-fossil fuel power production.

[ 2 ]

(i)

Outline the energy transformations that take place in an OWC.

[ 2 ]

(c)

(b) (i) An OWC design has an aperture that accepts a wave width of 4.5 m . The waves at the proposed site have an average wavelength of 95 m and wave period of 8.0 s . The overall efficiency of the energy conversion of the OWC is 24 %.

(not to scale)

Assuming that the waves have a rectangular cross-section, determine the minimum wave amplitude that will be required in order for the OWC to produce a power output of 0.10 MW .
Density of water $=1000 \mathrm{~kg} \mathrm{~m}^{-3}$

[ 3 ]

$\frac{P_{\mathrm{required}}}{w}=\frac{0.10\,\mathrm{MW}}{0.24\times4.5\,\mathrm{m}}=0.093\,\mathrm{MW\,m^{-1}}$ $v=\frac{\lambda}{T}=\frac{95\,\mathrm{m}}{8.0\,\mathrm{s}}=11.9\,\mathrm{m\,s^{-1}}$ $\frac{P}{w}=\frac12A^2\rho gv\quad\Longrightarrow\quad A=1.3\,\mathrm{m}$

(d)

On the grid, sketch a labelled Sankey diagram that represents the energy transformation in this OWC.

[ 3 ]

Diagram correct shape (but scaled incorrectly) and labelled.
Waste energy $0.32\ \text{MW}$ (allow ECF from (b)(i)).
All dimensions scaled correctly (by eye).

Figure for Answer to Question (d) — IB Physics SL

All question bank results loaded

IB Physics SL A 3 Work Energy and Power Topic Practice

Learn this topic

Question 1

Question (a)

Question (b)

Question (i)

Question 2

Question (a)

Question (b)

Question (i)

Question (ii)

Question (c)

Question 3

Question (a)

Question (i)

Question (ii)

Question (b)

Question (i)

Question (c)

Question (d)