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Transcript
```High School
by SSL Technologies
Part 3 /3
Physics Ex-38
Question-1
A force of 12 N, acting 60o from the horizontal, is applied to a 20 kg cart
initially at rest resulting in a final velocity of 10 m/s. If the force of friction
is 2 N, answer the following questions concerning the cart in going from
point-A to point-B (a distance of 250 m).
FA = 12 N
vi = 0
60o
A
FH = 6 N
vf = 10 m/s
f=2N
s = 250 m
20 kg
B
No “velocity not constant”
a) Was the cart at rest?
(While traveling from point-A to point-B)
12 N [E 60o N]
b) What was the applied force?
c) What was the horizontal component
of the applied force?
6 N right
d) What was the frictional force?
2 N left
e) What was the resultant force?
4 N right
(FR = FA – f = 6 N – 2 N = 4 N)
Click
Physics Ex-38
Question-1
A force of 12 N, acting 60o from the horizontal, is applied to a 20 kg cart
initially at rest resulting in a final velocity of 10 m/s. If the force of friction
is 2 N, answer the following questions concerning the cart in going from
point-A to point-B (a distance of 250 m).
FA = 12 N
vi = 0
60o
A
FH = 6 N
vf = 10 m/s
f=2N
s = 250 m
20 kg
B
f) What was the acceleration of the cart? 0.2 m/s2
g) What was the initial EK of the cart?
h) What was the final EK of the cart?
0
1 000 J
i) How much work was done on the cart? 1000 J
j) What becomes of the work done on the cart?
It is transferred to the cart in the form of EK (faster speed).
Click
Physics Ex-38
Question-1
A force of 12 N, acting 60o from the horizontal, is applied to a 20 kg cart
initially at rest resulting in a final velocity of 10 m/s. If the force of friction
is 2 N, answer the following questions concerning the cart in going from
point-A to point-B (a distance of 250 m).
FA = 12 N
vi = 0
60o
A
FH = 6 N
f=2N
s = 250 m
vf = 10 m/s
20 kg
B
k) How much work was done to overcome friction?
l) What was the total work done?
500 J
1500 J
n) Summarize the amounts of work done:
1000 J (1500 J – 500 J)
(Cart was not raised)
0
3) To overcome
500 be
J parallel
Remember
that friction
the force must
4)to
Total
done in the Work
1500Formula.
J
the work
distance
1) To accelerate the cart
2) To raise the cart
Click
Physics Ex-38
Question-2
A force of 110 N, acting 60o from the horizontal, is applied to a 10 kg cart whose
initial velocity is 4 m/s. The final velocity is 16 m/s. If the frictional force of 5 N,
answer the following questions concerning the cart in going from point-A to
point-B (a distance of 24 m).
FA = 110 N
vi = 4 m/s
60o
A
FH = 55 N
vf = 16 m/s
f=5N
s = 24 m
10 kg
B
No “velocity not constant”
a) Was the cart at rest?
(While traveling from point-A to point-B)
110 N [E 60o N]
b) What was the applied force?
c) What was the horizontal component
of the applied force?
55 N right
d) What was the frictional force?
5 N left
e) What was the resultant force?
50 N right
(FR = FA – f = 55 N – 5 N = 50 N)
Click
Physics Ex-38
Question-2
A force of 110 N, acting 60o from the horizontal, is applied to a 10 kg cart whose
initial velocity is 4 m/s. The final velocity is 16 m/s. If the frictional force of 5 N,
answer the following questions concerning the cart in going from point-A to
point-B (a distance of 24 m).
FA = 110 N
vi = 4 m/s
60o
A
FH = 55 N
vf = 16 m/s
f=5N
s = 24 m
10 kg
B
f) What was the acceleration of the cart?
g) What was the initial EK of the cart?
5 m/s2
80 J
h) What was the final EK of the cart?
1 280 J
i) How much work was done on the cart?
1 200 J
Work done on cart = total energy – energy lost to friction
1280 J – 80 J = 1200 J
Click
Physics Ex-38
Question-2
A force of 110 N, acting 60o from the horizontal, is applied to a 10 kg cart whose
initial velocity is 4 m/s. The final velocity is 16 m/s. If the frictional force of 5 N,
answer the following questions concerning the cart in going from point-A to
point-B (a distance of 24 m).
FA = 110 N
vi = 4 m/s
f=5N
60o
A
vf = 16 m/s
FH = 55 N
10 kg
s = 24 m
B
j) How much work was done to overcome friction?
k) What was the total work done?
120 J
1320 J
l) Summarize the amounts of work done:
1) To accelerate the cart
2) To raise the cart
3) To overcome friction
4) Total work done
1200 J
0
120 J
1320 J
(1320 J – 120 J)
(Cart was not raised)
Click
Physics Ex-38
Question-3
A 10 kg cart is traveling at 12 m/s towards the right. If a frictional force
of 40 N stops the cart in a distance of 18 m, answer the following
questions concerning the cart in going from point-A to point-B.
vi = 12 m/s
A
vf = 0
f = 40 N
10 kg
s = 18 m
B
a) Was the cart at rest?
No “velocity not constant”
(While traveling from point-A to point-B)
b) What was the frictional force?
40 N left (or – 40 N)
c) What was the resultant force?
40 N left (or - 40 N)
d) What was the acceleration?
- 4 m/s2
Click
Physics Ex-38
Question-3
A 10 kg cart is traveling at 12 m/s towards the right. If a frictional force
of 40 N stops the cart in a distance of 18 m, answer the following
questions concerning the cart in going from point-A to point-B.
vi = 12 m/s
A
vf = 0
f = 40 N
10 kg
s = 18 m
B
e) What was the initial EK of the cart?
f) What was the final EK of the cart?
g) How much energy did the cart lose?
720 J
0
720 J
h) What becomes of the energy lost by the cart?
Used to overcome friction (lost as heat and sound)
Click Click
Question-4
Physics Ex-38
A hammer falls from a scaffold and 1.5 s later strikes
the ground with a kinetic energy of 157.5 J.
What is the weight of the hammer?
Click
Question-5
Physics Ex-38
A projectile, whose mass is 800 g, is shot into the air with a
velocity of 25 m/s, 42o N of E. Determine the kinetic energy
of the projectile one second after it is fired.
Click
Question-6
Physics Ex-38
Starting from rest, a car reaches a velocity of 60 m/s in a distance
of 120 m. Assuming the system is frictionless and knowing that
the motor of the car produces a force of 3 x 104 N, calculate the
mass of the car.
Click
Question-7
Physics Ex-38
The mass of an electron is 1.67 x 1027 kg. What work must be done
on the electron in order to give it a speed of 2.5 x 107 m/s?
Click
Question-8
Physics Ex-38
A bullet of mass 2 g, traveling at 500 m/s, is fired at a piece of wood.
The bullet emerges from the wood with a speed of 100 m/s. If the
retarding force of friction was 4800 N, calculate the thickness of the
piece of wood. ?
Click
Question-9
Physics Ex-38
What is the mass of a stone that is thrown in the air with a
velocity of 1.1 m/s and with an initial kinetic energy of 0.0121 J?
Click
Question-10
Physics Ex-38
Two vehicles, X and Y, are traveling at the same speed.
Vehicle-X has twice the kinetic energy of vehicle-Y.
What is the value of the following ratio?
Mass of vehicle-X
Mass of vehicle-Y
Click
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```