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Tuesday, October 13, 2009

Gravitational potential energy to kinetic energy and vice versa.

 

As we have seen in earlier post an object can have either gravitational potential energy or kinetic energy. However it is always possible for the object to have both kinetic energy and gravitational potential energy. Think of a plane flying at a certain height above the ground.

It is going to have kinetic energy due to its speed and gravitational potential energy due to its height.

Now what happens to a body that is either falling toward the ground or rising to a certain height. 

body moving up and down

As you can see in fig 1 the object is rising and as a result its height is also rising, hence its potential energy is also increasing. Its kinetic energy is however decreasing. (see Principle of conservation of energy).

However the object in fig 2 is falling towards the ground. Since its height is decreasing, its gravitational potential energy must also be decreasing and its kinetic energy increasing.

Example 1

potential to kinetic energy

In this example a ball is released from rest from a height h.

If the ball is initially at rest then

v = 0  ms-1 hence Ek = 0 J

However since the object is at a height h the Ep = mgh

When the ball is released it fall under the effect of the force of gravity, as a result it will accelerate downward and as a result the velocity of the object will increase  while the height of the object decreases.

Hence as the velocity decreases the kinetic energy decreases while the gravitational potential energy decreases as the height decreases.

However at all time    Ek +  Ep = Total energy and total energy is constant.

As the object reaches the ground the height becomes zero so does the gravitational potential energy while the kinetic energy reaches its maximum value. At this point all the gravitational potential energy would have been converted to kinetic energy.

When it reaches the ground h = 0 m hence Ep = 0 J

While Ek = 0.5 mv2

Example 1

A man of mass 64 kg jumps from a bridge 25 m high into a river.

(a) Calculate the gravitational potential energy of the man when he is on the bridge.

(b) What is his speed of entry into the water.

Now the man is on the bridge at a height of 25 m. It means that he has gravitational potential energy.

(a)  Gravitational potential energy Ep = mgh

                                                                = 64*9.81*25

                                                                = 15696J

                                                                 = 1.5 *104J                                        

(b) When the person jumps the gravitational potential energy decreases as his height decreases. However as the person fall to the ground his speed increases and as a result the kinetic energy is completely converted to kinetic energy.

Hence what he reaches the river all the gravitational potential energy has been converted to kinetic energy.

Kinetic energy at surface of river =  1.5 *104J

Ek = 0.5 mv2

1.5 *104= 0.5 *64*v2

v = (1.5 *104/0.5/64)0.5

    = 21.65 m s-1

=22 m s-1

It is now time for a question. I will give you an answer to do. I will give the answer when some of you have given the answers.

Good luck.

A girl of mass 50 kg is trying to jump over a bar. She ran at a speed of and leaves the ground and successfully jumped over the bar.

(a) Calculate the kinetic energy that she has when she is running.

(b) Deduce the gravitational potential energy of the girl when she is at her maximum height.

(c) Calculate the height of the bar.

Monday, October 5, 2009

How to determine uncertainty in derived quantity when multiplication or division is performed

This is the second part of a series of post on finding uncertainty in derived quantities. You can find the index here and the part on addition and subtraction here.

Now very often when you performed an experiment you meet quantities that need to be processed to obtain a derived quantity such as g, the acceleration due to gravity, or any other derived quantities.

You have already seen the first part on addition and subtraction and now you will see how to do it for multiplication and division.

If in an experiment is performed and the following quantities are measured with their uncertainties.

A = 10.2+-0.2 cm

B = 5.4 +-0.4 cm

Now if you need to process these quantities to find AB and A/B with their uncertainties how would you do it?

Uncertainty in multiplication

How would you determine the value of AB and its uncertainty?

You will have to calculate the value of AB first.

AB = 10.2 *5.4 = 55.o8 =55 (2 sf )

To determine the uncertainty in AB you will have to use the equation below.

formula for uncertainty multiplication

Rearranging the equation will give you

uncertainty in mutiplication

Hence AB = 55+-5 cm2

Remember the uncertainty in A and B are to 1 sf hence the uncertainty in AB must be given to 1 sf.

Uncertainty in division

You are now going to determine the value of A/B and its uncertainty.

You will have to determine the value of A/B first.

A/B = 10.2 /5.4 =1.888 = 1.9

To determine the uncertainty of A/B  you will determine the equation below

image

Rearranging the equation will give you

image

Remember the uncertainty in A/B is given to 1 sf since the uncertainties in A and B are given to 1 sf.

Hence A/B = 1.9 +-0.2

With these two formula you can thus determine the uncertainty in any derived quantities that involves multiplication and division.

Now you can move to the next part where the method to determine uncertainty of derived quantities where powers,  square root, etc are involved.

How to determine uncertainty in a derived quantity when addition or subtraction is performed?

 

As we have seen in the previous post it is important to know how to determine the uncertainty for derived quantities.

Let us see how to determine how to determine the uncertainty of a derived quantity when an addition or a subtraction is performed.

In an experiment two quantities are measured as shown.

A = 56.4 +_0.1 cm

B = 12.2+_0.1 cm

If a derived quantity C is given by the equation below

C = A + B

Then the uncertainty in C  deltaC = deltaA +deltaB

                    deltaC = 0.1 + 0.1

                       = o.2 cm

C = 56.4 + 12.2 = 68.6 cm

C = 68.6 +-0.2 cm

If you want to calculate C and the equation relating C to A and B is

C = A – B

Then we can use the following equation to calculate deltaC

deltaC = deltaA  +deltaB

Hence deltaC  = 0.1 + 0.1  = 0.2 cm

If C = 56.4 – 12.2 = 44.2 cm

then C = 44.2 +-0.2 cm

This is then end of this part. Please follow the link below to go to the part on determining the

Saturday, October 3, 2009

Uncertainty and how to process it!

In physics all quantity that is measured come with its uncertainty. This is because the instrument, the experiment or the person performing the measurement is subjected to some limitations.

For example

  1. Someone that is measuring a length with a metre rule cannot measure the length to a precision that is better than one mm. So a length that is measured with a metre rule will have an uncertainty of +-o.1 c m.
  2. The limitation can either be due to the experiment itself, the apparatus,  or the experimenter.  Someone measuring the height of rebound will measure it with an uncertainty of +- 0.5 cm. Someone estimating the mass of an object can measure it with an uncertainty of +-0.5 kg. A voltmeter will give a reading with a percentage uncertainty of 1%.

[As you may have guess from 2 above not all measurements that are performed will have these uncertainty. These are for example only.]

As you can see all quantities that are measured will have uncertainties. And as a result any other quantities that are derived from the measured quantities will have their uncertainties.

We are going to see how to determine the uncertainty when the following operations are performed.

  1. Addition and subtraction
  2. multiplication and division
  3. power, square root, square, etc
  4. Other operations using the extreme-value method (under construction)

To know how to determine the uncertainty when addition and subtraction is performed see the next part of this series.

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