First Name:
Second Exam PHYS 1114
1.
Last Name:
5:30–6:20 PM, March 7, 2017
a) What is Newton’s Second Law in formula form?
b) What does each letter (or symbol) stand for in this equation?
c) What are the corresponding SI units? (Write: symbol $ unit).
d) A 76 kg man pushes against a 200 kg box on the floor, that does
not move. The magnitude of the friction force on the box is 105 N
due south. Give a free-body diagram with all applicable forces on
the box.
e) If the man pushes in the horizontal direction, determine the force
(magnitude and direction) on the box by the man.
f) Finally, determine the acceleration (magnitude and direction) of
the man during his push, assuming that he slides without friction.
First Name:
Second Exam PHYS 1114
Last Name:
5:30–6:20 PM, March 7, 2017
2. A car of mass m goes with constant speed v over the top of a hill that
may be described by part of a vertical circle of radius r.
a) Give the formula for what the net force on the car must be to
follow that path and also give its direction.
b) Does the net force do work on the car? And why or why not?
c) Give the free-body diagram for the car at the top of the hill, when
the speed is critical, so that the wheels barely stay on the road
and the driver feels weightless.
d) Derive the formula for the critical speed vc .
e) Assume that the mass of the car with its driver is 1500 kg and that
the radius r = 30 m. Evaluate vc in km/hr.
First Name:
Second Exam PHYS 1114
Last Name:
5:30–6:20 PM, March 7, 2017
3. A block of mass 15 kg is at rest on an inclined plane that makes an
angle of 30.9 with respect to the horizontal. The coefficient of static
friction µs = 0.6, whereas the coefficient of kinetic friction µk = 0.4.
a) Draw the appropriate free-body diagram.
b) Calculate the magnitude of the static friction force and compare
it with its maximal value.
c) Because of a small tremor the block starts sliding down. What is
the magnitude of the friction force now?
d) What is the magnitude of the net force on the block after the
tremor?
e) How large is the magnitude of the acceleration experienced by the
block then?
f) What is the speed of the block as it has slid 2.5 m along the inclined
plane?
First Name:
Second Exam PHYS 1114
Last Name:
5:30–6:20 PM, March 7, 2017
4. A 1600 kg safe is lifted to a twelfth-story window 33.0 m above the
ground using a steam engine.
a) Give the formula for the work done by a force F on an object that
moves distance d along a path making angle ✓ with the direction
of the force.
b) What is the defining formula for power delivered?
c) What does each symbol in part (b) represent?
d) What are the SI units for force, work, and power?
e) What is the minimum power required to lift the safe within 2.00
minutes?
f) What is the change in potential energy of the safe? Is this equal
to the work done on the safe using the formula in part (a)?
First Name:
Last Name:
Second Exam PHYS 1114
1.
Answer Sheet
5:30–6:20 PM, March 7, 2017
a) What is Newton’s Second Law in formula form?
X
F = ma
b) What does each letter (or symbol) stand for in this equation?
X
= Sum,
F = Force,
m = Mass,
a = Acceleration
c) What are the corresponding SI units? (Write: symbol $ unit).
X
$ No units,
2
F $ Newton = N = kg·m/s ,
2
m $ kilogram = kg,
a $ m/s
d) A 76.0 kg man pushes against a 200 kg box on the floor, that does
not move. The magnitude of the friction force on the box is 105 N
due south. Give a free-body diagram with all applicable forces on
the box.
FN
North
Fpush
Ffr
South
mg
e) If the man pushes in the horizontal direction, determine the force
(magnitude and direction) on the box by the man.
105 N due north (cancels friction)
f) Finally, determine the acceleration (magnitude and direction) of
the man during his push, assuming that he slides without friction.
105 N
2
= 1.38 m/s
76 kg
2
or 1.4 m/s
due south (Third Law used)
First Name:
Last Name:
Second Exam PHYS 1114
Answer Sheet
5:30–6:20 PM, March 7, 2017
2. A car of mass m goes with constant speed v over the top of a hill that
may be described by part of a vertical circle of radius r.
a) Give the formula for what the net force on the car must be to
follow that path and also give its direction.
Fnet
mv 2
= Fc =
r
towards center of circle
b) Does the net force do work on the car? And why or why not?
No: Force is perpendicular to the (circular) path
c) Give the free-body diagram for the car at the top of the hill, when
the speed is critical, so that the wheels barely stay on the road
and the driver feels weightless.
mg = Fc
d) Derive the formula for the critical speed vc .
mv 2
mg =
r
=)
v 2 = gr
=)
vc =
p
gr
e) Assume that the mass of the car with its driver is 1500 kg and that
the radius r = 30 m. Evaluate vc in km/hr.
q
2
vc = (9.80 m/s )(30 m) = 17.15 m/s = 62 km/h
✓
17.15 m 3600 s 1 km
61.7 km
=
1s
1 h 1000 m
1h
◆
First Name:
Last Name:
Second Exam PHYS 1114
Answer Sheet
5:30–6:20 PM, March 7, 2017
3. A block of mass 15 kg is at rest on an inclined plane that makes an
angle of 30.9 with respect to the horizontal. The coefficient of static
friction µs = 0.6, whereas the coefficient of kinetic friction µk = 0.4.
a) Draw the appropriate free-body diagram.
FN
Ffr
q
mg sinq
mg cosq
mg
q
b) Calculate the magnitude of the static friction force and compare
it with its maximal value.
fs = mg sin ✓ = 75.49 N
! fs,max = µs FN = µs mg cos ✓ = 75.68 N
c) Because of a small tremor the block starts sliding down. What is
the magnitude of the friction force now?
fk = µk FN = µk mg cos ✓ = 50.454 N = 50.5 N
d) What is the magnitude of the net force on the block after the
tremor?
Fnet = mg sin ✓
fk = 75.491 N
50.454 N = 25.04 N = 25.0 N
e) How large is the magnitude of the acceleration experienced by the
block then?
a=
Fnet
25.04 N
2
2
=
= 1.669 m/s = 1.7 m/s
m
15 kg
f) What is the speed of the block as it has slid 2.5 m along the inclined
plane?
q
2
2
v = 2ad =) v = 2(1.669 m/s )(2.5 m) = 2.889 m/s = 2.9 m/s
First Name:
Last Name:
Second Exam PHYS 1114
Answer Sheet
5:30–6:20 PM, March 7, 2017
4. A 1600 kg safe is lifted to a twelfth-story window 33.0 m above the
ground using a steam engine.
a) Give the formula for the work done by a force F on an object that
moves distance d along a path making angle ✓ with the direction
of the force.
W = F d cos ✓
b) What is the defining formula for power delivered?
P =
W
t
c) What does each symbol in part (b) represent?
P = Power,
W = Work Done,
t = Elapsed Time
d) What are the SI units for force, work, and power?
Force: N, Work: J=Nm,
Power: W=J/s
e) What is the minimum power required to lift the safe within 2.00
minutes?
F = mg ) W = mgh )
2
mgh
1, 600 kg ⇥ 9.80 m/s ⇥ 33.0 m
P =
=
= 4, 310 W
t
120 s
f) What is the change in potential energy of the safe? Is this equal
to the work done on the safe using the formula in part (a)?
PE = mgh = mg ⇥ h = W,
so yes!
Extra: The work delivered by the steam engine is
2
W = (1600 kg)(9.80 m/s )(33.0 m) = 517, 440 J = 517 kJ