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Physics Ex-39

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Potential energy is stored energy because it has the “potential” of being used at a future time. The potential energy of an object, EP ,depends upon its position.

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When an object is raised, work is done. The energy used to raisethe object is in the form of gravitational potential energy or justsimply the potential energy. The formula for potential energy isas follows:

EP = mgh

where: EP = the Potential Energy (in joules) m = the mass (in kilograms) g = the acceleration due to gravity 9.8 m/s2 or (10 m/s2) h = the height (in meters)

Also, since w = mg

ClickEP = wh

Note that for any object,EP is directly proportional

to the height.

Physics Ex-39

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In raising an object, the potential energy gained by the object does not depend upon the path of the object. That is, the heightis the perpendicular distance from the horizontal. In theillustration below, a 1 kg object is raised 1 metre. In each case,the potential energy gained by the object is 10 J.

PART-1

IMPORTANTThe potential energy of a system depends upon where

we choose the base for the height, h. W

Physics Ex-39

The Law of Conservation of energy states that energy cannot becreated nor destroyed. Thus, when an object is raised, work is doneand the object stores the energy in the form of potential energy.

When the object falls freely, the potential energy in converted intokinetic energy. Gradually, as the object falls down, the loss inpotential energy becomes the gain in kinetic energy. However,at all points during the fall, the total energy is the sum of thepotential energy plus the kinetic energy and remains constant.

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PART-1

EP = MaximumEK = 0

EK = MaximumEP = 0

ET = EP + EK

Physics Ex-39

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PART-1

In solving energy problems In solving energy problems involving the inclined plane, involving the inclined plane,

separate the energy calculationsseparate the energy calculationsinto three parts as listed below.into three parts as listed below.

HINT

Work to overcome friction: W = fs (If the system is ideal, then skip this part)

Work to accelerate the object: EK

(Consider the inclined plane as an ideal horizontal plane)

Work to raise the object: EP or W = wg = mgh

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Use g = 10 m/s2 for the Earth’s gravitational

acceleration.

Reminder

Physics Ex-39Question-1

A 20 kg object is raised 3 metres. Calculate the work done andtell where the energy went.

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The work goes to the object in the form of potential energy (EP).

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A ball is thrown up in the air. Explain the change in energyof the ball while going up and while coming back down.

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a) Going up: ______________________________________________

b) Coming down: ______________________________________________

The ball loses EK but gains in EP.

The ball loses EP but gains in EK.


A 1400 kg car is traveling at 10 m/s. Upon arriving at a hill,the car is allowed to coast. How high up the hill will the cartrise before coming to a stop?

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An empty swing is at its highest point 3 mfrom the ground and at its lowest point 1 mfrom the ground. What is its maximumspeed at its lowest point?

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3 m – 1 m


An object is thrown vertically upward. Which graph representsthe potential energy of the object as a function of its height?

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A 10 kg object falls from a height of 12 m. Fill in the potential,kinetic and total energy of the object at the given points.

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1 200 J

01 200 J

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900 J

300 J1 200 J

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600 J

600 J1 200 J

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300 J

900 J1 200 J

0

1 200 J1 200 J

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A 1 kg mass is fired into the air with a vertical velocity of 30 m/s.Fill in the potential, kinetic and total energy of the object for eachsecond it rises.

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450 J0 450 J

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400 J

50 J450 J

250 J

250 J450 J

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0

450 J450 J

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A 2 kg object falls from rest. Fill in the potential, kinetic andtotal energy of the object for the first 3 seconds of fall.

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900 J0 900 J

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800 J100 J 900 J

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800 J100 J 900 J

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0900 J 900 J

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An object starts from rest and slides down a frictionless ramp froma height of 10 m. What is the speed of the object at the bottom ofthe ramp?

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Which of the following graphs correctly illustrates the relationshipbetween the kinetic energy of a car versus its velocity.

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A 20 kg block is pushed up an incline at a constant velocityof 6 m/s by a force applied parallel to the incline (FA).

As illustrated in the diagram, the inclineis 10 m long and 5 m high. Assumingthe system is frictionless, answerthe following questionsconcerning the block whilesliding up the incline.

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Part-A: Work done to accelerate the block.

a) What is the change in velocity? (v)

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0

b) What is the acceleration?

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0

c) What is the initial EK of the block?

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360 J

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d) What is the final EK of the block?

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360 J

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e) How much EK did the car gain?

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0

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Physics Ex-39

h) How much work is done to raise the block?

Question-11

A 20 kg block is pushed up an incline at a constant velocityof 6 m/s by a force applied parallel to the incline (FA).

As illustrated in the diagram, the inclineis 10 m long and 5 m high. Assumingthe system is frictionless, answerthe following questionsconcerning the block whilesliding up the incline.

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Part-B: Work done to raise the block (relative to base line).

f) What is the weight of the block?

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200 N

g) What height is the block raised?

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5 m

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1 000 J

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i) What is the initial EP of the block?

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0

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j) What is the final EP of the block?

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1 000 J

ClickClickClickk) How much EP did the block gain? 1 000 J

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“Given”


A 20 kg block is pushed up an incline at a constant velocity of 6 m/s by a force applied parallel to the incline (FA).

As illustrated in the diagram, the inclineis 10 m long and 5 m high.

If the force of friction is 20 N,answer the following questionsconcerning the block whilesliding up the incline.

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0


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0


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360 J

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360 J

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e) How much EK did the block gain?

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0

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200 N


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5 m

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1 000 J

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0

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1 000 J

ClickClickClickk) How much EP did the block gain? 1 000 J

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“Given”

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n) What is the total work done?

Part-C: Work done to overcome friction.

l) What is the frictional force?

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20 N

m) What work is done to overcome friction?

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200 J

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1 200 J

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friction

“given ”

WT = Increase in EK + Increase in EP + Work to overcome friction

= 0 + 1000 J + 200 J = 1200 J


Starting from rest, a 20 kg block is pushed 10 m up anincline resulting in a final velocity of 10 m/s.

As illustrated, the force applied (FA) actsparallel to the incline therebyraising the block 5 m.

Assuming there is no friction,answer the following questionsconcerning the block whilesliding up the incline.

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10 m/s


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0


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360 J

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360 J

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e) How much EK did the block gain?

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0

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200 N


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5 m


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1 000 J

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0

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1 000 J

ClickClickk) How much EP did the block gain? 1 000 J

“ given ”

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Physics Ex-39

WT = Increase in EK + Increase in EP + Work to overcome friction

= 100 J + 1000 J + 0 = 2000 J

Question-13




Part-C: Work done to overcome friction.

f) What is the frictional force? 0

g) What work is done to overcome friction? 0

h) What is the total work done? 2 000 J

“ given ”

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“ frictionless ”

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Starting from rest, a 20 kg block is pushed 10 mup an incline resulting in a final velocity of 10 m/s.

As illustrated, the force applied (FA) actsparallel to the incline thereby raisingthe block 5 m.

If the force of friction is 20 N,answer the followingquestions concerning theblock while sliding up the incline.

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10 m/s


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0


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360 J

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360 J

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e) How much EK did the car gain?

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0

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Work done to accelerate the mass.


A tool whose mass is 600 g, falls 12 m into a box of sand.If the tool sinks 4 cm into the sand, calculate the (average)stopping force of the sand.

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