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PHYSICS

InClass

by SSL Technologies

with S. Lancione

Exercise-38

Kinetic Energy

Part 3 /3

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

vf = 10 m/s

60o

f = 2 N

20 kg

FH = 6 N

B

A

s = 250 m

No

a)

Was the cart at rest?

“velocity not constant”

(While traveling from point-A to point-B)

12 N [E 60o N]

b)

What was the applied force?

6 N right

c)

What was the horizontal component

of the applied force?

2 N left

d)

What was the frictional force?

4 N right

e)

What was the resultant force?

(FR = FA – f = 6 N – 2 N = 4 N)

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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

vf = 10 m/s

60o

f = 2 N

20 kg

FH = 6 N

B

A

s = 250 m

0.2 m/s2

f)

What was the acceleration of the cart?

0

g)

What was the initial EK of the cart?

1000 J

h)

What was the final EK of the cart?

1000 J

i)

How much work was done on the cart?

j)

What becomes of the work done on the cart?

It is transferred to the cart in the form of EK (faster speed).

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Remember that the force must be parallel

to the distance in the Work Formula.

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

vf = 10 m/s

60o

f = 2 N

20 kg

FH = 6 N

B

A

s = 250 m

500 J

k)

How much work was done to overcome friction?

1500 J

l)

What was the total work done?

n)

Summarize the amounts of work done:

1000 J

(1500 J – 500 J)

1) To accelerate the cart

(Cart was not raised)

0

2) To raise the cart

500 J

3) To overcome friction

4) Total work done

1500 J

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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

vf = 16 m/s

60o

f = 5 N

10 kg

FH = 55 N

B

A

s = 24 m

No

a)

Was the cart at rest?

“velocity not constant”

(While traveling from point-A to point-B)

110 N [E 60o N]

b)

What was the applied force?

55 N right

c)

What was the horizontal component

of the applied force?

5 N left

d)

What was the frictional force?

50 N right

e)

What was the resultant force?

(FR = FA – f = 55 N – 5 N = 50 N)

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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

vf = 16 m/s

60o

f = 5 N

10 kg

FH = 55 N

B

A

s = 24 m

5 m/s2

f)

What was the acceleration of the cart?

80 J

g)

What was the initial EK of the cart?

1280 J

h)

What was the final EK of the cart?

1200 J

i)

How much work was done on the cart?

Work done on cart = total energy – energy lost to friction

1280 J – 80 J = 1200 J

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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

vf = 16 m/s

60o

f = 5 N

10 kg

FH = 55 N

B

A

s = 24 m

120 J

j)

How much work was done to overcome friction?

1320 J

k)

What was the total work done?

l)

Summarize the amounts of work done:

1200 J

(1320 J – 120 J)

1) To accelerate the cart

(Cart was not raised)

0

2) To raise the cart

120 J

3) To overcome friction

4) Total work done

1320 J

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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

vf = 0

f = 40 N

10 kg

B

A

s = 18 m

No

a)

Was the cart at rest?

“velocity not constant”

(While traveling from point-A to point-B)

40 N left (or – 40 N)

b)

What was the frictional force?

40 N left (or - 40 N)

c)

What was the resultant force?

-4 m/s2

d)

What was the acceleration?

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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

vf = 0

f = 40 N

10 kg

B

A

s = 18 m

720 J

e)

What was the initial EK of the cart?

0

f)

What was the final EK of the cart?

720 J

g)

How much energy did the cart lose?

h)

What becomes of the energy lost by the cart?

Used to overcome friction (lost as heat and sound)

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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?

Answer

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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 secondafter it is fired.

Answer

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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.

Answer

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The mass of an electron is 1.67 x 1027 kg. What work must be done

on the electron in order to give it a speed of 2.5 x 107 m/s?

Answer

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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. ?

Answer

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What is the mass of a stone that is thrown in the air with avelocity of 1.1 m/s and with an initial kinetic energy of 0.0121 J?

Answer

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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

Answer

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