Work energy principle and power

In summary: The initial horizontal speed is the same as the final speed.The cosine of the initial angle is the horizontal speed divided by the total initial speed.So we can use the final horizontal speed to find the initial angle.Can you show how you got there?Sure. Let's start with the equation for conservation of energy in the horizontal direction:$\text{Total energy} = mgh + \frac 12 mv^2 = mgh + \frac 12 mv_x^2 + \frac 12 mv_y^2$Since the vertical speed is 0 at the final point, we can simplify the equation to:$\text{Total energy} = mgh + \frac 12 mv_x^2$Now let's plug
  • #1
Shah 72
MHB
274
0
A football is kicked from the ground level with speed 15m/s and rises to a height of 1.45m. Assume that air resistance is negligible
a) Find the speed of the ball when it is 1m above the ground
Increase in GPE= loss of KE
mgh=1/2mu^2-1/2mv2
10×(1.45-1)= 1/2×15^2-v^2/2
V= 14.7 m/ s
But textbook ans is 14.3m/s

Show that the ball was kicked at an angle of 21.0 degree
10× 1.45 sin theta= 1/2× 15^2- 1/2× 14.7^2
I don't get the ans.
 
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  • #2
The speeds and heights are mixed up.
Initially we have $v_0=15$ and $h_0=0$.
At $h_1=1\,\text{m}$, we have the unknown $v_1$ that we want to find.
And at $h_2=1.45\,\text{m}$, we have another unknown $v_2$, although we do know that the vertical speed is $v_{y,2}=0$ at this point.

Consider that the horizontal speed $v_x$ is constant.
In this case we can actually apply conservation of energy separately for both the horizontal and the vertical direction.
That is, at all times we have $\text{Total energy} =mhg + \frac 12 mv^2 = mhg + \frac 12 m(v_x^2 + v_y^2) = mhg + \frac 12 m v_x^2 + \frac 12 m v_y^2$, where $v_x$ is constant. Consequently the term with $v_x$ is the same each time, and we can remove it from the both sides of the corresponding conservation of energy equations.

Can we find the formulas for the total initial energy, the total energy at $h_1=1\,\text{m}$, and the total energy at $h_2=1.45\,\text{m}$, and fill in what we can?
 
Last edited:
  • #3
Klaas van Aarsen said:
The speeds and heights are mixed up.
Initially we have $v_0=15$ and $h_0=0$.
At $h_1=\,\text{m}1$, we have the unknown $v_1$ that we want to find.
And at $h_2=1.45\,\text{m}$, we have another unknown $v_2$, although we do know that the vertical speed is $v_{y,2}=0$ at this point.

Consider that the horizontal speed $v_x$ is constant.
In this case we can actually apply conservation of energy separately for both the horizontal and the vertical direction.
That is, at all times we have $\text{Total energy} =mhg + \frac 12 mv^2 = mhg + \frac 12 m(v_x^2 + v_y^2) = mhg + \frac 12 m v_x^2 + \frac 12 m v_y^2$, where $v_x$ is constant. Consequently the term with $v_x$ is the same each time, and we can remove it from the both sides of the corresponding conservation of energy equations.

Can we find the formulas for the total initial energy, the total energy at $h_1=1\,\text{m}$, and the total energy at $h_2=1.45\,\text{m}$, and fill in what we can?
So I get v= 14.7m/s
 
  • #4
Shah 72 said:
So I get v= 14.7m/s
That answer was deduced from $mgh_2 - mgh_1=\frac 12mv_0^2 - \frac 12 m v_1^2$, which is invalid, because $\frac 12mv_2^2$ is missing.

Instead we should have $mgh_0+\frac12 mv_0^2=mgh_1+\frac12 mv_1^2=mgh_2+\frac12 mv_2^2$, and we should also use that $v_x^2+v_y^2=v^2$ with constant $v_x$.
 
  • #5
So according to what you said I get the ans 15.3m/s
 
  • #6
Shah 72 said:
So according to what you said I get the ans 15.3m/s
Can you show how you got there?
 
  • #7
I u
Klaas van Aarsen said:
That answer was deduced from $mgh_2 - mgh_1=\frac 12mv_0^2 - \frac 12 m v_1^2$, which is invalid, because $\frac 12mv_2^2$ is missing.

Instead we should have $mgh_0+\frac12 mv_0^2=mgh_1+\frac12 mv_1^2=mgh_2+\frac12 mv_2^2$, and we should also use that $v_x^2+v_y^2=v^2$ with constant $v_x$.
I understood. Thanks a lott!
 
  • #8
Klaas van Aarsen said:
Can you show how you got there?
Initial GPE= mgh= 0J
Initial KE= 225/2J
Middle GPE= mgh= 10J
Middle KE= 1/2 mv^2
Final GPE= 14.5J
Final KE = 0J
So I calculated for Initial and Middle one
0+ 225/2=10+v^2/2
So I get v=14.3m /s when h=1m
 
  • #9
Klaas van Aarsen said:
Can you show how you got there?
Can you also pls pls guide me with the next question where I have to prove that the ball is kicked at an angle of 21.0 degrees with the horizontal.
 
  • #10
Shah 72 said:
Final KE = 0J
The final KE is not 0. The velocity still has the constant horizontal speed component $v_x$.
So the final KE is $\frac 12mv_x^2$.
 
  • #11
Shah 72 said:
Can you also pls pls guide me with the next question where I have to prove that the ball is kicked at an angle of 21.0 degrees with the horizontal.
If we calculate the constant $v_x$, we can find the initial angle.
 
  • #12
Klaas van Aarsen said:
If we calculate the constant $v_x$, we can find the initial angle.
So you mean I have to calculate the horizontal distance?
Can you pls help me
 
  • #13
Klaas van Aarsen said:
The final KE is not 0. The velocity still has the constant horizontal speed component $v_x$.
So the final KE is $\frac 12mv_x^2$.
Let me try to work it out
 
  • #14
Klaas van Aarsen said:
If we calculate the constant $v_x$, we can find the initial angle.
So final velocity = 13.6m/s
How do I find the angle?
 
  • #15
Shah 72 said:
So final velocity = 13.6m/s
How do I find the angle?
The initial horizontal speed is the same as the final speed.
The cosine of the initial angle is the horizontal speed divided by the total initial speed.
 
  • #16
Klaas van Aarsen said:
The initial horizontal speed is the same as the final speed.
The cosine of the initial angle is the horizontal speed divided by the total initial speed.
Iam finding it a bit difficult. Can you pls show me a little bit of working
 
  • #17
Let $\theta$ be the initial angle.
Then according to the definition of the cosine, we have
$$\cos\theta=\frac{v_x}{v_0}$$
The initial speed is given as $v_0=15$ and you have just calculated the final speed, which is horizontal, and the same as the initial horizontal speed $v_x$.
 
  • #18
Klaas van Aarsen said:
Let $\theta$ be the initial angle.
Then according to the definition of the cosine, we have
$$\cos\theta=\frac{v_x}{v_0}$$
The initial speed is given as $v_0=15$ and you have just calculated the final speed, which is horizontal, and the same as the initial horizontal speed $v_x$.
Thank you so so so so so so so much!
 

FAQ: Work energy principle and power

What is the work-energy principle?

The work-energy principle states that the work done on an object is equal to the change in its kinetic energy. This means that when a force is applied to an object, energy is transferred to the object and it gains kinetic energy.

How is work calculated?

Work is calculated by multiplying the force applied to an object by the distance the object moves in the direction of the force. This can be represented by the equation W = Fd, where W is work, F is force, and d is distance.

What is power?

Power is the rate at which work is done or energy is transferred. It is calculated by dividing the amount of work done by the time it takes to do the work. The unit of power is watts (W).

How is power related to work and energy?

Power is directly related to work and energy. The more work is done or energy is transferred in a given amount of time, the higher the power output. This means that a greater amount of power is required to do more work or transfer more energy in a shorter period of time.

How is the work-energy principle applied in real-life situations?

The work-energy principle is applied in many real-life situations, such as when a person lifts a weight or when a car accelerates. In these situations, work is done on the object, transferring energy to it and causing a change in its kinetic energy. This principle is also used in machines, where a small amount of force is applied over a longer distance to do a larger amount of work in a shorter period of time.

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