Topic: work, energy and power Far-north schools

Class test Date: May 2021

Physical sciences Total = 30 marks

Question 1
Multiple choice questions

Four options are provided as possible answers to the following questions.

Each question has only ONE correct answer.

1.1. An applied force F accelerates an object of mass m on a horizontal

frictionless surface from a velocity v to a velocity 2v.

The net work done on the object is equal to ...

1 2
A mv
2
2
B mv
3 2
C mv
2
2
D 2mv (2)

1.2. Two balls, P and Q, are dropped simultaneously from the same height. Ball
P has TWICE the mass of ball Q. Ignore the effects of air friction.
Just before the balls hit the ground, the kinetic energy of ball P is x. The
kinetic energy of ball Q, in terms of x, will be …

1
A x
4
1
B x
2
C x
D 2x (2)

1.3. The speed of a bicycle increases from 2 m∙s-1 to 8 m∙s-1. Its kinetic energy
increases by a factor of …
 A 4
B 6
C 8
D 16 (2)

Question 2

The diagram below shows a track, ABC. The curved section, AB, is frictionless. The
rough horizontal section, BC, is 8 m long.

An object of mass 10 kg is released from point A which is 4 m above the

ground. It slides down the track and comes to rest at point C.

2.1.1 State the principle of conservation of mechanical energy in words. (2)

2.1.2 Using ENERGY PRINCIPLES only, calculate the magnitude of the frictional
force exerted on the object as it moves along BC. (5)

2.2 A motor pulls a crate of mass 300 kg with a constant force by means of a
light inextensible rope running over a light frictionless pulley as shown
below. The coefficient of kinetic friction between the crate and the surface
of the inclined plane is 0,19.

2.2.1 Calculate the magnitude of the frictional force acting between the crate and
the surface of the inclined plane. (3)

The crate moves up the incline at a constant speed of 0,5 m∙s-1.

2.2.2 Calculate the average power delivered by the motor while pulling the crate
up the incline. (5)
[15]
 QUESTION 3

A rescue helicopter is stationary (hovers) above a soldier. The soldier of mass 80 kg is

lifted vertically upwards through a height of 20 m by a cable at a CONSTANT SPEED
of 4 m∙s-1. The tension in the cable is 960 N. Assume that there is no sideways motion
during the lift. Air friction is not to be ignored.

960 N

80 kg

3.1 State the work-energy theorem in words. (2)

3.2 Draw a labelled free-body diagram showing ALL the forces acting on the
soldier while being lifted upwards. (3)

3.3 Use the WORK-ENERGY THEOREM to calculate the work done on the
soldier by friction after moving through the height of 20 m. (4)
[09]

TABLE 2: FORMULAE

MOTION
v f =v i +a Δt Δx=v i Δt+ 12 aΔt 2 Δy =v i Δt+ 12 aΔt 2
or/of

v 2= v 2 + 2 aΔx
f i or/of
v 2= v 2 + 2 aΔy
f i Δx= ( vi + vf
2 ) Δt
or/of
Δy = ( vi + vf
2 ) Δt

FORCE/KRAG
F net =ma p =mv
F net Δt = Δp
w = mg
Δp= mv f − mv i
WORK, ENERGY AND POWER
W = FΔx cosθ U = mgh or/of
E P = mgh
1 1 W net =ΔK W net =ΔE k
K= mv 2 E k= mv 2 or/of
2 or/of 2
ΔK =K f −K i or/of
ΔE k =E kf −E ki
W
P= P = Fv
Δt
 Work MEMO

Question 1
Multiple choice questions

1.1. C 
1.2. B 
1.3. D  (6)

Question 2

2.1.1 In an isolated/closed system,  the total mechanical energy is conserved /

remains constant  (2)

OR
The total mechanical energy of a system is conserved/ remains constant
 in the absence of friction. 

2.1.2 Along AB
Wnet = ΔEk 
FGΔhcosθ = ½ m(vf2 – vi2)
(10)(9,8)(4)cos0o = ½ (10)(vf2 – 0) 
v = 8,85 m∙s-1
f
Substitute 8,85 m∙s-1 in one of the following options

Along BC or Along BC
Wnet = ΔK Wnc = ΔK + ΔU 
fΔx cosθ = ΔK f Δxcosθ = ΔK + ΔU
f(8)cos 180o = ½ (10)(0 – 8,852)  f(8)cos 180 = ½ (10)(0 - 8,852) + 0
f = 48,95 N f = 48,95 N (Accept 49 N)
(5)
2.1.3 fk = μkN
= μkmgcosθ
= (0,19)(300)(9,8) cos 25o
= 506,26 N (3)

2.1.4 OPTION 1

Fnet = 0
Fapp + (-FGsinθ) + (-f) = 0
Fapp - (300)(9,8)sin 25o - 506,26 = 0
Fapp = 1 748,76 N
Pave = Fvave 
= 1748,76 x 0,5
= 874,38 W  (5)

OPTION 2
Wf + Wapp + WN + Wg = 0
FΔxcosθ + FappΔx cosθ + 0 + FgΔx cosθ = 0
(506,26Δx cos180o)  + (FappΔx cos0) + 300(9,8)Δxcos115o= 0
Fapp = 1748,76 N
Pave = Fvave
= (1748,76 ) (0,5) 
= 874,38 W
QUESTION 3

3.1 The net work done on an object is equal to 

the change in kinetic energy of the object.  (2)

Accepted Labels
Fg / Fw /force of Earth on victim/weight / 1 470 N / mg / gravitational force
w
Fapplied F / force on victim / 1 600 N / FA

Ffriction / Ff / friction / air resistance

f

 Accept : Force diagram

3.2 
Fapplied
Fapplied


f
f
w 
w

(3)

3.3 Option 1/Opsie 1: Notes

Wnet = ∆K  OR IF Wnet = ∆K not
WF + Ww + Wf = ∆K used:
F∆ycos  + Fw∆ycos  + Wf = ∆K Fnet = F + Fw + Ff
(960)(20)cos0° + (80)(9,8)(20)cos180°  + Wf = 0 0 = 960 – 784 + Ff
19 200 – 15 680 + Wf = 0  Ff = - 176 N
Wf = - 3 520 J  Wf = Ff∆ycos 
Option 2: = (176)(20)cos180°
Wnet = ∆K  OR WF - ∆U + Wf = ∆K
F∆ycos  - (mghf – mghi) + Wf = ∆K IF
(960)(20)cos0° - [(80)(9,8)(20) – 0]  + Wf = 0 Wnet = ∆K 
19 200 - 15 680 + Wf = 0  Fnet∆ycos  = ∆K
Wf = - 3 520 J  

(4)
[09]