Problem#1
A 0.600-kg particle has a speed of 2.00 m/s at point A and kinetic energy of 7.50 J at point B. What is (a) its kinetic energy at A? (b) its speed at B? (c) the total work done on the particle as it moves from A to B?
Answer:
(a) Kinetic energy at A is
KA = ½ mvA2 = ½ (0.600 kg)(2.00 m/s)2 = 1.20 J
(b) speed at B is
KB = ½ mvB2 = 7.50 J
½ (0.600 kg)vB2 = 7.50 J
vB = 5.00 m/s
(c) the total work done on the particle as it moves from A to B is
∑W = ∆K = KB – KA
∑W = 7.50 J – 1.20 J = 6.30 J
Problem#2
A 0.300-kg ball has a speed of 15.0 m/s. (a) What is its kinetic energy? (b) What If? If its speed were doubled, what would be its kinetic energy?
Answer:
(a) kinetic energy ball given by
K = ½ mv2 = ½ (0.300 kg)(15.0 m/s)2 = 33.8 J
(b) If its speed were doubled, kinetic energy ball is
K = ½ (0.300 kg)(30.0 m/s)2 = 135 J
Problem#3
A 3.00-kg object has a velocity (6.00i – 2.00j) m/s. (a) What is its kinetic energy at this time? (b) Find the total work done on the object if its velocity changes to (8.00i + 4.00j) m/s. (Note: From the definition of the dot product, v2 = v.v)
Answer:
Given: vi = (6.00i – 2.00j) m/s
(a) vi = {vix2 + viy2}1/2 = {(6.00 m/s)2 + (–2.00 m/s)2} = √40 m/s
Ki = ½ mvi2 = ½ (3.00 kg)(√40 m/s)2 = 60.0 J
(b) vf = (8.00i + 4.00j) m/s
vf2 = vf.vf = 64.0 + 16.0 = 80.0 m2/s2
∆K = Kf – Ki = ½ m(vf2 – vi2)
∆K = ½ (3.00 kg)(80.0 m2/s2) – 60 J = 60.0 J
Problem#4
A 2 100-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 5.00 m before coming into contact with the top of the beam, and it drives the beam 12.0 cm farther into the ground before coming to rest. Using energy considerations, calculate the average force the beam exerts on the pile driver while the pile driver is brought to rest.
Answer:
Consider the work done on the pile driver from the time it starts from rest until it comes to rest at
the end of the fall. Let d = 5.00 m represent the distance over which the driver falls freely, and
h = 0.12 m the distance it moves the piling.
∑W = ∆K
Wgravity + Wbeam = ½ mvf2 – ½ mvi2, so
mg(h + d) cos00 + (Favg)d cos1800 = 0 – 0
Thus,
Favg = mg(h + d)/d
Favg = (2100 kg)(9.80 m/s2)(5.12 m)/0,120 m = 8.78 x 105 N.
The force on the pile driver is upward.
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