SPH3U1 March 29, 2016 – Forces

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SPH3U1
March 29, 2016
Chapter 3 Test – Forces
ANSWERS
Part 1 – Multiple Choice
1.
2.
3.
4.
B
B
B
E
Knowledge and Understanding – 12 marks
5.
6.
7.
8.
E
C
B
C
9.
10.
11.
12.
Part 2 – Diagrams
A
C
B
A
Communication – 9 marks
1. The graph has a constant positive slope. As mass is decreasing at a constant rate, so
acceleration will increase at the same constant rate.
π‘Ž (π‘š/𝑠 2 )
𝑑 (𝑠)
2. The net force on Block A is to the right. The net force acting on the system is not.
Block A accelerates at the same rate as the system, however it has a smaller mass. Therefore,
the net force acting on Block A must have a smaller magnitude than that acting on the system.
3. The net force acting on this object is 2.0 N [R]. Therefore, the object will accelerate to the right.
In other words, the velocity will change, increasing to the right.
Part 3 – Problem Solving
Thinking & Investigation – 18 marks
1. The maximum force will be equal to the weight of the woman and boxes.
𝐹𝑔 = π‘šπ‘šπ‘Žπ‘₯ 𝑔
(950)
= π‘šπ‘šπ‘Žπ‘₯
(9.8)
96.9 π‘˜π‘” = π‘šπ‘šπ‘Žπ‘₯
π‘šπ‘šπ‘Žπ‘₯ = π‘šπ‘€π‘œπ‘šπ‘Žπ‘› + π‘šπ‘π‘œπ‘₯𝑒𝑠
(96.9) − (85) = π‘šπ‘π‘œπ‘₯𝑒𝑠
11.9 π‘˜π‘” = π‘šπ‘π‘œπ‘₯𝑒𝑠
11.9
= 2.83
4.2
∴ π‘‘β„Žπ‘’ π‘€π‘œπ‘šπ‘Žπ‘› π‘π‘Žπ‘› π‘π‘Žπ‘Ÿπ‘Ÿπ‘¦ 2 π‘π‘œπ‘₯𝑒𝑠.
SPH3U1
Wednesday April 12, 2011
2. 𝑣𝑓 = 𝑣𝑖 + π‘ŽΔ𝑑
(18.6) = (0.0) + π‘Ž(5.1)
(18.6)
=π‘Ž
(5.1)
3.65 π‘š/𝑠 2 = π‘Ž
𝐹𝑁𝐸𝑇 = π‘šπ‘Ž
𝐹𝑁𝐸𝑇 = (1480)(3.65)
𝐹𝑁𝐸𝑇 = 5402
∴ π‘‘β„Žπ‘’ 𝑛𝑒𝑑 π‘“π‘œπ‘Ÿπ‘π‘’ 𝑖𝑠 π‘Žπ‘π‘Ÿπ‘œπ‘₯π‘–π‘šπ‘Žπ‘‘π‘’π‘™π‘¦ 5400 𝑁
3.
a. Block B accelerates with the system, so aB = aS
𝐹𝑁𝐸𝑇𝑆 = π‘šπ‘† π‘Žπ‘†
(8.0) = (6.6 + 9.5)π‘Žπ‘†
8.0
= π‘Žπ‘†
6.6 + 9.5
0.497 = π‘Žπ‘†
∴ π΅π‘™π‘œπ‘π‘˜ 𝐡 π‘Žπ‘π‘π‘’π‘™π‘’π‘Ÿπ‘Žπ‘‘π‘’π‘  π‘Žπ‘‘ π‘Žπ‘π‘π‘Ÿπ‘œπ‘₯π‘–π‘šπ‘Žπ‘‘π‘’π‘™π‘¦ 0.50 π‘š/𝑠 2
b. The force of tension is responsible for accelerating Block A, which also accelerates at the
same rate as the system.
𝐹𝑁𝐸𝑇𝐴 = π‘šπ΄ π‘Žπ΄
𝐹𝑇 = π‘šπ΄ π‘Žπ΄
𝐹𝑇 = (6.6)(0.497)
𝐹𝑇 = 3.28
∴ π‘‘β„Žπ‘’ π‘‘π‘’π‘›π‘ π‘–π‘œπ‘› 𝑖𝑛 π‘‘β„Žπ‘’ π‘ π‘‘π‘Ÿπ‘–π‘›π‘” 𝑖𝑠 π‘Žπ‘π‘π‘Ÿπ‘œπ‘₯π‘–π‘šπ‘Žπ‘‘π‘’π‘™π‘¦ 3.3 𝑁
4. 𝐹𝑁𝐸𝑇 = π‘šπ‘Ž
(75.0) = (340)π‘Ž
(75.0)
=π‘Ž
(340)
0.221 π‘š/𝑠 2 = π‘Ž
𝑣𝑓2 = 𝑣𝑖2 + 2π‘ŽΔ𝑑
SPH3U1
Wednesday April 12, 2011
𝑣𝑓 = √𝑣𝑖2 + 2π‘ŽΔ𝑑
𝑣𝑓 = √(0.0)2 + 2(0.221)(48)
𝑣𝑓 = 4.61
∴ π‘‘β„Žπ‘’ π‘“π‘–π‘›π‘Žπ‘™ 𝑠𝑝𝑒𝑒𝑑 𝑖𝑠 π‘Žπ‘π‘π‘Ÿπ‘œπ‘₯π‘–π‘šπ‘Žπ‘‘π‘’π‘™π‘¦ 4.6 π‘š/𝑠
5. 𝐹𝑁𝐸𝑇 = π‘šπ‘Ž
𝐹𝑁𝐸𝑇 = πΉπ‘Žπ‘π‘ − 𝐹𝑔
πΉπ‘Žπ‘π‘ − 𝐹𝑔 = π‘šπ‘Ž
πΉπ‘Žπ‘π‘ = π‘šπ‘Ž + 𝐹𝑔
πΉπ‘Žπ‘π‘ = (5.0)(1.4) + (5.0)(9.8)
πΉπ‘Žπ‘π‘ = 56
∴ π‘‘β„Žπ‘’ π‘π‘Ÿπ‘Žπ‘›π‘’ π‘Žπ‘π‘π‘™π‘–π‘’π‘  π‘Ž π‘“π‘œπ‘Ÿπ‘π‘’ π‘œπ‘“ 56 𝑁
Part 4 – Short Answer
Application – 10 marks
1. Answers may vary.
2. “Action at a distance” forces such as the force of gravity, magnetic force, and electric force can all
accelerate objects without contact.
3. According to Newton’s First Law, the object must be experiencing uniform motion. It may be
moving at a constant velocity.
4. Answers may vary.
ο‚· Weight is the force of gravity. In space, the force of gravity is present, so objects still
have weight
ο‚· Humans cannot perceive the force of gravity, so this feeling of weightlessness is the
absence of the normal force
ο‚· The normal force is absent (or diminished) because the environment (i.e. space capsule)
is accelerating with the person.
ο‚· This relates to Newton’s 3rd Law because the normal force is a “reaction” force that
results when the force of gravity pulls an object down onto on a stationary surface.
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