**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?**

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The couch above is in equilibrium : Rotational equilibrium and translational equilibrium.

Taking moments about the left point,

Anti-clockwise moments about left point = Clockwise moments about right point

F_{By right person}(2) = F_{Weight of person on couch}(0.5) + F_{Weight of couch}(1)

F_{By right person}(2) = (60)(9.8)(0.5) + (20)(9.8)(1)

F_{By right person} = 490/2

F_{By right person} = 245 N

Taking moments about the right point,

Clockwise moments about right point = Anti-clockwise moments about left point

F_{By left person}(2) = F_{Weight of person on couch}(1.5) + F_{Weight of couch}(1)

F_{By left person}(2) = (60)(9.8)(1.5) + (20)(9.8)(1)

F_{By left person} = 539 N

Answer: 245N and 539N

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

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

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

Answer: 3

**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?**

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Let

m

_{1}be the 5.00 kg object

m

_{2}be the 10.00 kg object

m

_{f}be the 15.0 kg object(result of the collision)Using conservation of momentum,

m

_{1}u

_{1}+ m

_{2}u

_{2}= m

_{f}v

5.00(15.0) + 10.0(0) = 15.0(v)

v = 5.00 ms

_{-1}Loss of kinetic energy = ½m

_{1}u

_{1}

^{2}– ½m

_{f}v

^{2}

= ½(5.00)(15.0)

^{2}– ½(15.0)(5.00)

^{2}

= 375 J

Answer: 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.**

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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^{-1}ii) 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?**

**2.0 kW****20 kW****4.0 kW****5.0 kW**

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Power output = mgh/t

Efficiency = output/input

input = output /efficiency

= 200 x (8/4) / 0.20

= 2000 W

Answer: 1

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