Download Chapter 3 Problem Set

Chapter 3 Problem Set
2. How much work is needed to raise a 100-kg load of bricks 12 m above the ground to a
building under construction?
Answer:
The relevant formulas here are:
W = F X d (work = force X distance) and
F = m X a (force = mass X acceleration)
In this case we are give the mass (110-kg), and the distance (12 m), and we know the
acceleration due to gravity is 9.8 m/sec2. So we first solve for the force and then solve for the
work done.
F = m X a = 110 kg X 9.8 m/sec2 = 1,078 kg*m/sec2 = 1,078 N (newtons)
Plugging that into the formula for work:
W = F X d = 1,078 N X 12 m = 12,936 N*m = 12,936 J (joules)
10. What is the speed of an 800-kg car whose KE (kinetic energy) is 250 kJ?
The relevant formula here is:
KE = ½ mv2 (kinetic energy = ½ X mass X velocity2
In this case we are give the mass (800-kg) and the kinetic energy (250 kJ) and we are asked
to solve for the speed of the car. So first we convert kilojoules (kJ) into joules :
250 kJ = 250 kJ X 1000 J/kJ = 250,000J
Rearanging the formula for kinetic energy to solve for velocity we get:
NOTE: The symbol  followed by brackets [ ] will be used to show the square root
function for all problems. So the square root of 25 would be shown as [25].
V2 = [2 X (KE)/m]
v = [(2 X 250000 J)/800kg] = [(2 X 250000 N*m)/800 kg]
= [(2 X 250000 kg*m2/sec2)/800 kg]
= [(500,000 k*m2/sec2)/800 kg]
= [625 m2/sec2] = 25 m/sec
15. A 70-kg athlete runs up the stairs from the ground floor of the Empire State Building to its
one hundred second floor, a height of 370 m, in 25 min. How much power did the athlete
develop?
The relevant formulas here are:
F = m X a (force = mass X acceleration; here the acceleration is that generated by gravity g)
W = F X d (work = force X distance),
P = W/t (power = work/time)
We are given the athlete’s mass (70-kg), and the distance involved is the height that he runs
to (370 m) and we know that the acceleration due to gravity is 9.8 m/sec2. We are also give
the amount of time required to do the amount of work the athlete did (25 min).
First, solving for the force involved:
F = mg = 70 kg X 9.8 m/sec2 = 686 kg*m/sec2 = 686 N (newtons)
Now solving for the work done:
W = F X d = mgh = 686 N X 370 m = 253,820 N*m = 253,820 J (joules)
Before we can solve for power we have to convert the time (25 min) into seconds:
t = 25 min X 60 sec/min = 1,500 sec
Now solving for the power:
P = W/t = 253,820 J/1500 sec = 169 w (watts)
24. A boy throws a 4-kg pumpkin at 8 m/sec to a 40-kg girl on roller skates, who catches it.
At what speed does the girl then move backward?
This is a case of the law of the conservation of momentum. The relevant formula is:
p = m X v (momentum = mass X velocity)
We are given the mass (4-kg) and velocity (8 m/sec) of the pumpkin so we can calculate its
momentum (p). We are also given the weight of the girl (40-kg). Since momentum is
conserved, we can rearrange the formula (p = m X v) and calculate for the speed of the girl.
Solving for the momentum of the pumpkin:
p = m X v = 4 kg X 8 m/sec = 32 (kg*m)/sec
Rearranging the equation to solve for the girl’s velocity:
v = p/m = (velocity = momentum/mass) = [32 (kg*m)/sec]/40 kg = 0.8 m/s
31. Approximately 5.4 X 106 J of chemical energy is released when 1 kg of dynamite
explodes. What fraction of the total energy of the dynamite is this?
The relevant formula here is:
E0 = mc2 [resting energy = mass X (speed of light)2]
We are given the mass of the dynamite (1-kg), the chemical energy (Ec) given off by the
explosion (5.4 X 106 J) and we know the speed of light (3 X 108 m/sec) so we can calculate
the total resting energy (E0) of the dynamite:
E0 = mc2 = 1 kg X (3 X 108 m/sec)2 = 1 kg X (9 X 1016 m2/sec2) = 9 X 1016 kg*m2/sec2
= 9 X 1016 J (joules)
The fraction of the total energy given off as chemical energy is a ratio of the two:
Ef = Ec/E0 (fractional energy = chemical energy/total resting energy)
Ef = (5.4 X 106 J)/(9 X 1016 J) = 6 X 10-11 or 6 X 10-9 % (0.000000006 %)