# Questions for Forces and Dynamics (JC) Set 4

Two people are carrying a 2 m long couch with a mass of M=20kg level with the horizontal by lifting it from it ends. The center of mass of the couch is 1 m from each end. While it is being carried, another person with a mass m of 60 kg sits on the couch 0.5 m from one end. How much force will each person carrying the couch have to exert upwards in order to continue carrying the couch levelly? The couch above is in equilibrium : Rotational equilibrium and translational equilibrium.

Taking moments about the left point,
FBy right person(2) = FWeight of person on couch(0.5) + FWeight of couch(1)
FBy right person(2) = (60)(9.8)(0.5) + (20)(9.8)(1)
FBy right person = 490/2
FBy right person = 245 N

Taking moments about the right point,
FBy left person(2) = FWeight of person on couch(1.5) + FWeight of couch(1)
FBy left person(2) = (60)(9.8)(1.5) + (20)(9.8)(1)
FBy left person = 539 N

A submarine is in equilibrium in a fully submerged position. What causes the upthrust on the submarine?

1. The air exerts a greater upward force on the submarine than the weight of the steel.
2. The air in the submarine is less dense than the sea water.
3. There is a difference in water pressure acting on the top and bottom of the submarine.
4. The submarine displaces its own volume of water.

Upthrust occurs because the pressure in a fluid increases with increasing depth.

A 5.00 kg object moves at 15.0 ms-1. It collides perfectly inelastically with a 10.0 kg object which was at rest. How much kinetic energy is lost in the collision?

Let
m1 be the 5.00 kg object
m2 be the 10.00 kg object
mf be the 15.0 kg object(result of the collision)Using conservation of momentum,
m1u1 + m2u2 = mfv
5.00(15.0) + 10.0(0) = 15.0(v)
v = 5.00 ms-1Loss of kinetic energy = ½m1u12 – ½mfv2
= ½(5.00)(15.0)2 – ½(15.0)(5.00)2
= 375 J

a) State Newton’s second law of motion.

b) A 80.0 kg man is parachuting and experiencing a downward acceleration of 2.5 m s-2. The mass of the parachute is 10.0 kg. Calculate the upward force exerted on the system (parachute and man) by the air.

c) Upon reaching the ground, the man uses a pellet gun and fires one pellet per second with a speed of 483 m s-1 perpendicularly onto a wall. The pellet is stopped by the wall and the mass of each pellet is 2.14 g.
i) Find the initial momentum of each pellet.

ii) If each pellet is in contact with the wall for 1.25 ms, calculate the average force exerted on the wall by each pellet while in contact.

iii) Discuss qualitatively, what happens to the average force exerted on the wall by each pellet while in contact if the contact time is increased.

a) Newton’s second law of motion states that the rate of change of momentum of a body is directly proportional to the net external force on the body, and the change in momentum is in the direction of the force.b) By Newton’s second law of motion,
W – F = ma
(80 + 10)(9.81) – F = (80 + 10.0)(2.5)
F = (80 + 10.0)(9.81 – 2.5)
= 657.9
= 660 Nc)

i) Initial momentum = mv
= 2.14 x 10-3 x 483
= 1.0 kg m s-1ii) By Newton’s second law of motion,
force on each bullet by wall = [ (0 – 1.033) / (1.25 x 10-3)]
= -826.896 NBy Newton’s third law of motion,
force on wall by each pellet = 826.896 N
= 830 N

iii) According to Newton’s second law of motion, if contact time is increased, the average force exerted on the wall by each pellet while in contact will decrease.

A crane is used to raise a weight of 200 N at a constant speed through a vertical height of 8.0 m in 4.0 s.
The efficiency of the crane is 20%. What is the electrical power needed to be supplied to the crane?

1. 2.0 kW
2. 20 kW
3. 4.0 kW
4. 5.0 kW

Power output = mgh/t
Efficiency = output/input
input = output /efficiency
= 200 x (8/4) / 0.20
= 2000 W

Back To Forces And Dynamics ##### Mini Physics

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