Hwa Chong Institution
Hwa Chong Institution
Sec3 CSE (HP) Physics
Sec3 CSE (HP) Physics
Prepared by Mr Tan Kaiyuan
CAA 280409
Answers to Worksheet
Name: _____________________________________
Date: ______________________
Class: 3G1
Answer the following questions on a piece of foolscap.
1. Two forces of 15.0 N and 20.0 N are applied to a block of mass 0.30 kg in the directions
shown in Fig 1 below. Assuming there is no friction present. What is the acceleration of
the block?
0.30 kg
20.0 N
15.0 N
Taking left as negative (-ve),
By Newton’s Second Law of Motion,
FR = 20.0 N – 15.0 N = ma
5.0 N = (0.30 kg) a
a = 17 ms-2 ( 2 s.f.)
2. A car is travelling at 30 ms-1 on a level road. At this speed, the car has overcome a total
force of 600 N opposng the car.
a. Calculate the distance travelled by the car in 10 s.
v = s/t
s = vt = (30 ms-1)(10 s)
s = 300 m (s.f. may be 1, 2 or 3 since anything before decimal point is insignificant)
b. State the value of the driving force produced by the engine for a steady speed of 30
ms-1.
600 N
(constant velocity, hence no net forces, therefore driving force must be balanced by
opposing forces)
c. Explain why the car slows down when it climbs a hill even though the driving force is
unchanged.
This is because the NET FORCE acting on the vehicle is in the OPPOSITE direction
compared to that of the velocity of the vehicle. This results in the velocity decreasing.
(please turn to back for detailed explanation)
1
Prepared by Mr Tan Kaiyuan
CAA 280409
Hwa Chong Institution
Sec3 CSE (HP) Physics
N
Driving Force, FD = 600 N
WY = mg Cosθ
W = mg
WX = mg Sinθ
For those who are more adventurous, please look at the diagram above. The box
represents the vehicle on the slope. The weight, W, may be resolved into its two
components (the dotted arrows) along an x-axis and y-axis chosen by you. The
above x- and y-axes were chosen because they are convenient to work with. Once
the W is resovled into its components, it is completely replaced by them. It need not
be considered anymore. It is as though there are now four forces acting on the
vehicle (ie. N, FD, W X and W Y; W X may be shifted up and drawn on the vehicle),
instead of the original three (ie. N, FD, W). Because the vehicle does not move up or
down perpendicular to the slope, We may therefore infer that W Y and N cancels
each other out. Hence, we are left with W X and FD. If the vehicle slows down, there
must be a net force acting downhill. We may hence infer that W X > FD. Still cannot
understand? See attachment at the end of this answer key.
d. While travelling at 30 ms-1 on the level road, the driving force becomes zero. The
mass of the car is 800 kg. Calculate the deceleration of the car.
F = ma
(− 600 N) = (800 kg) a
a = − 0.75 ms-2 (s.f. not critical)
3. A box of mass 2.0 kg is pushed along a rough surface by an external force F of 10 N
such that it moves at constant speed.
a. Comment on the following statement:
Friction is both a friend and a foe.
Friend – Required for braking, required for walking so that you may not slip.
Foe – Reduces engine efficiency because energy is converted to heat which cannot
be used.
2
Prepared by Mr Tan Kaiyuan
CAA 280409
Hwa Chong Institution
Sec3 CSE (HP) Physics
b. Draw a free-body diagram of the 2.0 kg box, indicating its nature, line of action and
magnitude.
N
F
Ff
W
c. If F is increased to 20.0 N, calculate the resulting acceleration.
FR = 20.0 N – 10.0 N = ma
10.0 N = (2.0 kg) a
a = 5.0 ms-2 (2 s.f.)
4. A 3000 kg truck is connected to a 1000 kg trailer by a tow-bar. The whole system moves
at an acceleration of 2.0 ms-2. The friction acting on the truck is 800 N, and that on the
trailer is 200 N.
Tow-bar
FD
mtruck = 3000kg
Ffriction = 800 N
a. Calculate the driving force exerted by the engine of the truck.
Treating the whole system as a single body,
FR = FD – Ftotal friction = ma
FD – (800 N + 200 N) = (3000 kg + 1000 kg)(2.0 ms-2)
b. Calculate the tension acting on the tow-bar by the trailer.
On trailer only,
FR = T – Ff = ma
T = ma + Ff
T = (1000 kg)(2.0 ms-2) + 200 N
T = 2200 N
3
mtrailer = 1000kg
Ffriction = 200 N
Prepared by Mr Tan Kaiyuan
CAA 280409
Hwa Chong Institution
Sec3 CSE (HP) Physics
c. After some time, the whole system travels at a constant velocity of 10ms-1. State the
new value of the driving force.
1000 N (Constant velocity, ie. No net force acting on the whole system, hence
driving force must be equal to the frictional forces).
d. Describe the motion of the system when the engine is suddenly cut off in terms of its
velocity and acceleration.
The system continues to move forward with a decreasing velocity until it comes to
rest. The acceleration is throughout this motion is – 0.25 ms-2.
5. A 1200 kg lift accelerates uniformly from a velocity of 50 ms-1 to 60 ms-1 in 5 seconds.
a. Draw a free-body diagram representing the forces acting on the lift
If lift accelerating
downwards,
T
T
If lift accelerating
upwards,
W
W
.
b. Calculate the acceleration of the lift.
a = (v-u)/t
a = (60 – 50) / 5
a = 2 ms-2
c. Calculate the tension in the cable which holds the lift.
T – W = ma
T = ma + W
You may omit intermediate steps. I only wrote
T = ma + mg
these steps for those who cannot follow the math.
T = m (a + g)
T = 1200 kg ( 2 ms-2 + 10 ms-2)
T =14400 N
d. A man weighing 500 N enters the lift. Determine the magnitude of the force acting
on his feet.
The normal contact force of the lift acting on the man is 500 N, if the lift is not
accelerating. However, if the lift is accelerating upwards at 2 ms-2, then,
On the man,
N – W = ma
N = ma + mg = 50kg (2 ms-2) + 500 N = 600 N
4
Prepared by Mr Tan Kaiyuan
CAA 280409
Hwa Chong Institution
Sec3 CSE (HP) Physics
The normal contact force acting on his feet is 600 N. Likewise, he will also exert a
force of 600 N back on the lift. Hence, if you place a weighing scale in the lift, and
have him stand on it, the scale will read 600 N as well, because the weighing scale
measures the normal contact force of the man on the scale.
6. By means of a scale drawing, determine the resultant force acting on the block below:
10.0 N
30°
30°
10.0 N
10.0 N
10.0 N
30°
30°
30°
40.0 N
10.0 N
This diagram is not drawn to scale, however, if you
had drawn to scale, you should be able to obtain
40.0 N of the resultant force.
~end~
5
Prepared by Mr Tan Kaiyuan
CAA 280409
Hwa Chong Institution
Sec3 CSE (HP) Physics
Supplementary Notes
m = 10.0kg
N
30°
W
a) Determine the acceleration of the 10.0 kg box above down the frictionless
slope.
m = 10.0kg
30°
WY
By rearranging the free-body diagram a
little… shifting the Weight, W, so that it
may originate from the same point as the
Normal Contact Force, N, we obtain the
diagram on the left.
N
Let us resolve the Weight, W, into its Xand Y-components. Recall from vector
addition that Resultant Vectors are
summed from Component Vectors. We may
also reverse the process by resolving the
Resultant Vector into its components. Let
WX be the X-component of the weight that
is perpendicular to the surface of the
slope and WY be the Y-component of the
weight that is parallel to the surface of
30°
W
WX
the slope.
The Y-component of the Weight, WY, must be equal in magnitude as the Normal
Contact Force N. If it is not, the box will sink perpendicular to the slope surface
into the slope, or fly up! Since they are equal in magnitude, but pointing in opposite
directions, they cancel each other out. The only force that needs to be dealt with
is the X-component of the Weight, WX. (Note that the Weight no longer is of
concern because it has been resolved into its components; ie. It has been replaced
by the components).
WX is the only force left. In fact, it is the vector sum of N and W. It is thus the
resultant force in this example. Hence,
But
Sub
����⃗
������⃗
𝐹𝐹
⃗
𝑅𝑅 = 𝑊𝑊𝑋𝑋 = 𝑚𝑚𝑎𝑎
������⃗
���⃗
𝑊𝑊
𝑋𝑋 = 𝑊𝑊 sin 30°,
���⃗ = 𝑚𝑚𝑚𝑚
𝑊𝑊
���⃗ sin 30° = 𝑚𝑚𝑚𝑚
∴ 𝑊𝑊
∴ 𝑎𝑎 =
𝑚𝑚𝑚𝑚 sin 30° = 𝑚𝑚𝑚𝑚
𝑚𝑚𝑚𝑚 sin 30°
𝑚𝑚
6
= 𝑔𝑔 sin 30° = 5.0 ms-2