Questions for Forces and Dynamics (JC) Set 1


A spring A of force constant 6.0 Nm-1 is connected in series with a spring B of force constant 3.0 Nm-1. One end of the combination is securely anchored and a force of 0.60N is applied to the other end.
(a) By how much does each spring extend?
(b) What is the force constant of the combination?
(c) What is the total elastic potential energy stored in the system?

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(a)
F = Kaxa
0.6= 6(xa)
Extension for spring a, xa=0.100mF= kbxb
0.6=3(xb)
Extension for spring b, xb=0.200m

(b)
Total Extension = 0.10 +0.20
= 0.300m

F = kx
0.6 = k(0.3)
k = 2 Nm-1

(c)
Total elastic P.E. stored = (1/2)keffx2
= (1/2)(2)(0.3)2
= 0.090 J

 

A string supports a solid iron object of mass 0.180 kg totally immersed in a liquid of density 800 kg m-3. If the density of iron is 8000 kg m-3, calculate:
(a) the upthrust acting on the object;
(b) the tension in the string

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(a)
Upthrust = Vρg
= (0.180/8000)(800)(9.81)
= 0.177N(b)
Tension = (0.180)(9.81) – 0.177
= 1.59N

An ice cube of sides 2.0cm floats in a cup of tea. One of its faces is 0.20cm above the surface of the tea in the cup. Calculate the density of the tea if the density of ice is 920 kg m-3.

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Weight of ice = Weight of liquid displaced
Viceρiceg = Vteaρteag
(0.020)3(920) = (0.020)2(0.018)ρtea
ρtea=1.02 X 103 kg m-3

A parachutist of mass 80kg descends vertically at a constant velocity of 3.0 m s-1. Taking the acceleration of free fall as 10 ms-2, what is
(a) the net force acting on him?
(b) the upward force acting on him?

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(a) Net force is 0. Because the parachutist is at constant velocity. The gravitational force acting on him equals the resistance force acting on him(air resistance).(b)
m1g = F1
(80)(10) = F1
F1=800N

(a) A man throws a ball of mass 3.0kg with a speed of 5.0 ms-1. If his hand is in contact with the ball for a time interval of 0.20 s while throwing the ball, find the average force he exerts on the ball.
(b) If the man throws 4 balls in 2 seconds, find the average force exerted by him in one second.
(c) If the balls thrown by the man hit a wall and bounce back with speed 3.0 ms-1, find the average force exerted by the wall on the balls.

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(a)<F>= Δp/Δt
= (3)(5)/(0.20)
= 75N

(b)<F><t>= Δ in momentum in one collision x collision frequency
<F>(1) = 3.0 (5.0 – 0) X (4/2)
<F> = 30N

(c)
Δ in momentum of ball per collision = 3.0 (-3.0 – 5.0)
= -24 kg m s-1

<F> = -24 X (4/2)
= -48 N

Average force exerted by wall on ball is 48N

 

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