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Department of Natural Sciences
Clayton College & State University
October 2, 2002
Physics 1111 – Quiz 5
Name _______SOLUTION______________________________
A crate of 50.0 kg mass containing a new lab instrument is dragged by enthusiastic
physics students a distance of 30.0 m along a straight, level corridor to a physics lab.
The students maintain a constant pull of magnitude 200 N applied to the crate by means
of a rope inclined to the horizontal at angle of 40.0o. The velocity of the crate along the
straight path is not constant.
a. Calculate the work done by each force on the crate.
s = 30.0 m
A = 200 N
 = 40.0o
WA = A cos()s = (200.N) cos(40.0o) (30.0 m) = 4596 J
Wn = n cos()s = n cos(90.0o) (30.0 m) = 0 J
Ww = w cos()s = w cos(90.0o) (30.0 m) = 0 J
b. Calculate the total work done on the crate.
WNET = WA + Wn + Ww = 4596 J + 0 J + 0 J = 4596 J
c. If initially the crate was at rest, what is the final speed of the crate as it reaches
the lab?
WNET = ½ m Vf 2 – ½ mVi 2
Vi = 0 m/s
½ mVf 2 = WNET
Vf = sqrt( 2WNET/m) = 13.6 m/s
Department of Natural Sciences
Clayton College & State University
October 9, 2002
Physics 1111 – Quiz 6
Name _____SOLUTION________________________________
1. At the Longbranch saloon, a full beer stein of mass 2.00 kg is slid horizontally
(and without rotation) across the bar with an initial speed of 3.00 m/s. If the stein
travels a distance 1.53 m before stopping without loosing a drop of amber liquid,
find the coefficient of kinetic friction k for the stein on the bar surface.
Vi = 3.00 m/s
Vf = 0 m/s
KEi = ½ m Vi2 = ½ (2.00 kg)(3.00 m/s)2 = 9.00 J
KEf = 0 J
Work – Kinetic energy theorem including non-conservative forces:
Wnc = KEf - KEi = 0 J – 9.00 J = -9.00 J
Wnc = Wfriction = fk cos(180o)s = - fk s
m) = 5.88 N
fk = k n
k = fk /n
free-body diagram for the beer stein!)
fk = - Wnc / s = - (-9.00 J)/(1.53
n = mg (to prove it you need to draw a
k = fk /mg = (5.88 N)/ (2.00 kg x 9.80 m/s) = 0.300
Department of Natural Sciences
Clayton State University
October 13, 2005
Physics 1111 – Quiz 6.1
Name ____SOLUTION_________________________________
1. The units of work are
a. J.
b. N-m.
c. kg-m2/s2.
d. All of the above.
2. You throw a ball straight up. Compare the sign of the work done by gravity while the
ball goes up with the sign of the work done by gravity while it goes down.
a. Work is + on the way up and + on the way down.
b. Work is + on the way up and - on the way down.
c. Work is - on the way up and + on the way down.
d. Work is - on the way up and - on the way down.
3. If the net work done on an object is positive, then the object's kinetic energy
a. Decreases.
b. Remains the same.
c. Increases.
d. Is zero.
Department of Natural Sciences
Clayton State University
October 12, 2005
Physics 1111 – Quiz 6
Name ____SOLUTION_________________________________
1. If you push twice as hard against a stationary brick wall, the amount of work you do
a. Doubles.
b. Is cut in half.
c. Remains constant but non-zero.
d. Remains constant at zero.
2. You throw a ball straight up. Compare the sign of the work done by gravity while the
ball goes up with the sign of the work done by gravity while it goes down.
a. Work is + on the way up and + on the way down.
b. Work is + on the way up and - on the way down.
c. Work is - on the way up and + on the way down.
d. Work is - on the way up and - on the way down.
3. If the net work done on an object is negative, then the object's kinetic energy
a. Decreases.
b. Remains the same.
c. Increases.
d. Is zero.
Department of Natural Sciences
Clayton State University
October 9, 2007
Physics 1111 – Quiz 5
Name _SOLUTION____________________________________
A crate of 50.0 kg mass containing new physics textbooks is dragged by enthusiastic
physics students a distance of 30.0 m along a 25o incline to a physics lab.
The students maintain a constant pull of magnitude 250 N applied to the crate by means
of a rope parallel to the incline.
a. Calculate the work done by each force on the crate. Ignore friction.
WA = A x cos () = (250 N)(30.0m) = 7500 J
Ww = (mg) x cos () = (50.0 kg)(9.81 m/s2)(30.0m)cos () = -6219 J
Wn = n x cos () = 0
b. Calculate the total work done on the crate.
Wnet = WA + Ww + Wn = 1281 J
c. If initially the crate was at rest, what is the final speed of the crate as it reaches
the lab?
Wnet = Kf - Ki = ½ m Vf 2
Vf 2 = 2 Wnet /m
Vf = (2 Wnet /m)1/2
Vf = 7.16 m/s
Department of Natural Sciences
Clayton College & State University
February 28, 2005
Physics 1111 – Quiz 5
Name ____SOLUTION_________________________________
Farmer Benton hitches his tractor to a sled loaded with firewood and pulls it a
distance of 20.0 m along level ground. The total weight of the sled and load is 14,700 N.
The tractor exerts a constant 5000-N force at an angle of 36.9o to the horizontal, as shown
below. There is a 3500-N friction force opposing the sled’s motion. Find the work done
by each force acting on the sled and the total work done by all the forces.
Ww = w x cos (w) = (14,700 N) (20.0 m) cos (90o) = 0 J
Wn = n x cos (n) = (14,700 N) (20.0 m) cos (90o) = 0 J
WA = A x cos (A) = (5000 N) (20.0 m) cos (36.9o) = 79,968 J
Wf = fk x cos (f) = (3500) (20.0 m) cos (180o) = -70,000 J
Wtot = Ww + Wn + WA + Wf = 9,968 J
Department of Natural Sciences
Clayton State University
March 1, 2006
Physics 1111 – Quiz 5
Name ___SOLUTION__________________________________
1. An 88-g arrow is fired from a bow whose string exerts an average force of 110 N
on the arrow over a distance of 78 cm. What is the speed of the arrow as it leaves the
bow?
W = F x cos( = (110 N)(0.780 m)cos(0o) = 85.8 J
W = ½ m V2
V2 = 2 W/ m
V = (2 W/ m)1/2
V = (2 ()85.8 J/ (0.0880 kg))1/2 = 44.2 m/s
2. A 7.0-kg monkey swings from one branch to another 1.2 m higher. What is the
change in potential energy?
y = yf – yi = 1.20 m

U = m g yf – m g yi = m g (y) = (7.00 kg) (9.81 m/s2)(1.20 m) = 82.4 J
Department of Natural Sciences
Clayton State University
March 15, 2006
Physics 1111 – Quiz 6
Name __SOLUTION___________________________________
1.
A skier traveling 12.0 m s reaches the foot of a steady upward 18.0º incline and glides
12.2 m up along this slope before coming to rest. What was the average magnitude
of friction force acting on the skier?
Wnc = Ef - Ei
Ei = Ki + Ugi = ½ m Vi2
(Ugi = 0)
Ef = Kf + Ugf = m g yf
(Kf = 0)
Where yf = d sin () = (12.2 m) sin (18.0º) = 3.77 m
Wnc = Wfriction = fk cos(180o)s = - fk s
m) = 5.88 N
fk = - Wnc / s = - (-9.00 J)/(1.53
Department of Natural Sciences
Clayton College & State University
February 28, 2007
Physics 1111 – Quiz 6
Name _SOLUTION____________________________________
1. The quantity mgy is
a. The kinetic energy of the object.
b. The gravitational potential energy of the object.
c. The work done on the object by the force.
d. The power supplied to the object by the force.
2. An acorn falls from a tree. Compare its kinetic energy K, to its potential energy U.
a. K increases and U decreases.
b. K decreases and U decreases.
c. K increases and U increases.
d. K decreases and U increases.
3. A 30-N box is pulled 6.0 m up along a 37° inclined plane. What is the work done by
the weight (gravitational force) of the box?
a. - 11 J
b. - 1.1 × 102J
N)(6.00 m)(sin 37°) = -108 J
Wg = Ugi - Ugf = mgyi – mgyf = 0 – (30.0
c. - 1.4 × 102 J
d. - 1.8 × 102 J
Department of Natural Sciences
Clayton State University
February 27, 2008
Physics 1111 – Quiz 6
Name _____SOLUTION________________________________
You are pushing a 7.00 kg box up a rough wall with a force of 185 N directed at 30.0o to
the horizontal. The magnitude of the friction force acting on the box is 11.0 N.
a. Calculate the work done by each force on the box after you pushed it through
0.750 m.
W = F x cos ()
WA = A x cos() = (185.N) (0.750 m) cos(60.0o) = 69.4 J
Wn = n x cos() = n (0.750) cos(90.0o) = 0 J
Ww = wx cos() = (7.00 kg)(9.81 m/s2) (0.750 m) cos(180o)= - 51.5 J
Wf = fx cos() = (11.0 N)(0.750 m) cos(180o)= - 8.25 J
b. Calculate the total work done on the box.
WNET = WA + Wn + Ww + Wf = 69.4 J + 0 J -51.5 J - 8.25 J = 9.65 J
c. If initially the box was at rest, what is the final speed of the box after you
pushed it through 0.750 m?
WNET = ½ m Vf 2 – ½ mVi 2
Vi = 0 m/s
½ mVf 2 = WNET
Vf = sqrt( 2WNET/m) = 1.66 m/s
Department of Natural Sciences
Clayton State University
July 6, 2005
Physics 1111 – Quiz 7
Name __SOLUTION___________________________________
A 145-g baseball is dropped from a tree 13.0 m above the ground.
(a) With what speed would it hit the ground if air resistance could be ignored?
TEi = ½ mVi2 + m g yi = m g yi, since Vi = 0
TEf = ½ mVf2 + m g yf = ½ mVf2, since yf = 0
TEi = TEf
m g yi = ½ mVf2
g yi = ½ Vf2
Vf = (2 g yi )1/2
Vf = (2 (9.80 m/s2) (13.0 m) )1/2
Vf = 16. 0 m/s
(b) If it actually hits the ground with a speed of 8.00 m s , what is the average force of air resistance
exerted on it?
Wnet
= TEf - TEi = ½ mVf2- m g yi
Wnet = f x cos (180o) = -f x
-f x = ½ mVf2- m g yi
f = (- ½ mVf2 +m g yi)/x
f = 1.06 N
Department of Natural Sciences
Clayton State University
July 2, 2007
Physics 1111 – Quiz 7
Name __SOLUTION___________________________________
A skier with a mass of 70.0 kg glides certain distance up the slope (30.0o above the
horizontal) before coming to rest. If the speed of the skier at the bottom of the slope was
15.0 m/s and friction can be neglected, how far did he move along the slope?
Ei = m g yi + ½ mVi2 = ½ mVi2
corresponds to Ug = 0)
(We assume that the bottom of the slope
Ef = m g yf + ½ mVf2 = m g yf (since Vf = 0)
Ei = Ei
½ mVi2 = m g yf
½ Vi2 = g yf
(½ Vi2 )/g = yf
yf = 11.5 m
d = yf / sin(30.0o) = 23.0 m