# SPH3U1 March 29, 2016 – Forces

```SPH3U1
March 29, 2016
Chapter 3 Test – Forces
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 (π/π  2 )
π‘ (π )
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 &amp; Investigation – 18 marks
1. The maximum force will be equal to the weight of the woman and boxes.
πΉπ = ππππ₯ π
(950)
= ππππ₯
(9.8)
96.9 ππ = ππππ₯
ππππ₯ = ππ€ππππ + ππππ₯ππ
(96.9) − (85) = ππππ₯ππ
11.9 ππ = ππππ₯ππ
11.9
= 2.83
4.2
∴ π‘βπ π€ππππ πππ πππππ¦ 2 πππ₯ππ .
SPH3U1
Wednesday April 12, 2011
2. π£π = π£π + πΔπ‘
(18.6) = (0.0) + π(5.1)
(18.6)
=π
(5.1)
3.65 π/π  2 = π
πΉππΈπ = ππ
πΉππΈπ = (1480)(3.65)
πΉππΈπ = 5402
∴ π‘βπ πππ‘ πππππ ππ  πππππ₯ππππ‘πππ¦ 5400 π
3.
a. Block B accelerates with the system, so aB = aS
πΉππΈππ = ππ ππ
(8.0) = (6.6 + 9.5)ππ
8.0
= ππ
6.6 + 9.5
0.497 = ππ
∴ π΅ππππ π΅ πππππππππ‘ππ  ππ‘ ππππππ₯ππππ‘πππ¦ 0.50 π/π  2
b. The force of tension is responsible for accelerating Block A, which also accelerates at the
same rate as the system.
πΉππΈππ΄ = ππ΄ ππ΄
πΉπ = ππ΄ ππ΄
πΉπ = (6.6)(0.497)
πΉπ = 3.28
∴ π‘βπ π‘πππ πππ ππ π‘βπ π π‘ππππ ππ  ππππππ₯ππππ‘πππ¦ 3.3 π
4. πΉππΈπ = ππ
(75.0) = (340)π
(75.0)
=π
(340)
0.221 π/π  2 = π
π£π2 = π£π2 + 2πΔπ
SPH3U1
Wednesday April 12, 2011
π£π = √π£π2 + 2πΔπ
π£π = √(0.0)2 + 2(0.221)(48)
π£π = 4.61
∴ π‘βπ πππππ π ππππ ππ  ππππππ₯ππππ‘πππ¦ 4.6 π/π
5. πΉππΈπ = ππ
πΉππΈπ = πΉπππ − πΉπ
πΉπππ − πΉπ = ππ
πΉπππ = ππ + πΉπ
πΉπππ = (5.0)(1.4) + (5.0)(9.8)
πΉπππ = 56
∴ π‘βπ πππππ πππππππ  π πππππ ππ 56 π
Application – 10 marks
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.