Work energy principle and power

• MHB
• Shah 72
In summary, the golf ball of mass 45.9g is hit from a tee with speed 180km/ h. The ball rises to a height of 20m having traveled along a curved path of length 61.875m. At the highest point of its path the ball is traveling at 144 km/h. The average resistance remains unchanged. The ball travels a further 105.8m along a curved path to land on the green. The green is 4m lower than the tee. The speed of the ball just before it lands on the green is 35 m/s. The energy absorbed by the green is 30.3J.
Shah 72
MHB
A golf ball of mass 45.9g is hit from a tee with speed 180km/ h. The ball rises to a height of 20m having traveled along a curved path of length 61.875m. At the highest point of its path the ball is traveling at 144 km/h
a) Find the magnitude of the avg resistance force acting on the golf ball.
I got the and resistance =0.2N

The ball travels a further 105.8m along a curved path to land on the green. The green is 4m lower than the tee. The average resistance remains unchanged
b) find the speed of the ball just before it lands on the green
I got the ans v= 35m/s

The ball is traveling vertically when it lands on the green, where it is immediately brought to rest.
C) show that the energy absorbed by the green is 30.3J
I don't know how to work this out. Pls help

Shah 72 said:
I don't know how to work this out. Pls help
You don't know how to work what part out? Again, we can't really help you until we know what you are having troubles with! You have posted a number of problems without showing us anything about what you don't know. This has to end.

Let's start from the beginning. This problem is going to involve the work-energy theorem. You have been shown the form of this equation several times here so please start by telling us what the equation is and what your question is.

-Dan

topsquark said:
You don't know how to work what part out? Again, we can't really help you until we know what you are having troubles with! You have posted a number of problems without showing us anything about what you don't know. This has to end.

Let's start from the beginning. This problem is going to involve the work-energy theorem. You have been shown the form of this equation several times here so please start by telling us what the equation is and what your question is.

-Dan
It's question c iam not able to solve

Shah 72 said:
It's question c iam not able to solve
Since the ball didn't bounce (and we technically don't have to assume that) the green absorbed all the kinetic energy that the ball had when it landed. So that would be...

I ran through the whole problem and you are not off by much. I suspect that you are not carrying enough decimal places when you are rounding. Keep 5 or so significant digits. I got 0.1884 N for part a, 33.888 m/s for part b, and 26.356 J for part c. I couldn't replicate your results so I don't know where you might be off.

-Dan

Ok
topsquark said:
Since the ball didn't bounce (and we technically don't have to assume that) the green absorbed all the kinetic energy that the ball had when it landed. So that would be...

I ran through the whole problem and you are not off by much. I suspect that you are not carrying enough decimal places when you are rounding. Keep 5 or so significant digits. I got 0.1884 N for part a, 33.888 m/s for part b, and 26.356 J for part c. I couldn't replicate your results so I don't know where you might be off.

-Dan
Thank you so much for replying.
According to textbook the ans in 0.2N for q(a)
So for q(a)
m= 0.0459kg, u= 50m/s, h=20m, s=61.875m, highest point v=40m/s
So Increase in KE= 1/2mv^2- 1/2mu^2=-20.655J
Increase in GPE= mgh=9.18J
Work done against resistance = -F×S=-61.875F
So I get F= 0.2N as given in the textbook
For (b) s=105.8, h= 20-4= 16m, u=50m/s
Increase in KE=0.02295v^2-57.375J
Increase in GPE= 7.344J
Work done against resistance =-21.16J
So I get v=35.4m/s, as mentioned in the textbook I rounded it to 35 m/s.
I don't understand how to calculate q(c). I don't know what components to consider.

I
topsquark said:
Since the ball didn't bounce (and we technically don't have to assume that) the green absorbed all the kinetic energy that the ball had when it landed. So that would be...

I ran through the whole problem and you are not off by much. I suspect that you are not carrying enough decimal places when you are rounding. Keep 5 or so significant digits. I got 0.1884 N for part a, 33.888 m/s for part b, and 26.356 J for part c. I couldn't replicate your results so I don't know where you might be off.

-Dan
If I consider the final KE the ball lands = -28.11J
Final GPE= -1.836 J by taking h=4m
Work done against resistance = -0.2×4=-0.8J
So total will be 30.7. I don't get 30.3J .is this OK?

topsquark said:
Since the ball didn't bounce (and we technically don't have to assume that) the green absorbed all the kinetic energy that the ball had when it landed. So that would be...

I ran through the whole problem and you are not off by much. I suspect that you are not carrying enough decimal places when you are rounding. Keep 5 or so significant digits. I got 0.1884 N for part a, 33.888 m/s for part b, and 26.356 J for part c. I couldn't replicate your results so I don't know where you might be off.

-Dan
Can you pls pls explain an easy method how to calculate the energy gained by the green when the ball falls. If I get the concept right, I can calculate with similar problems. Iam not understanding which energy to take whether it's initial energy or its the final energy when the ball hits the green. Pls

I'm not sure what you are asking for. The best I can come up with is to say that if the ball lands and does not bounce then all of it's kinetic energy is absorbed by the green. That's simply a matter of energy conservation.

-Dan

topsquark said:
I'm not sure what you are asking for. The best I can come up with is to say that if the ball lands and does not bounce then all of it's kinetic energy is absorbed by the green. That's simply a matter of energy conservation.

-Dan
Ok got it. Thank you so much!

1. What is the work energy principle?

The work energy principle is a fundamental concept in physics that states that the work done on an object is equal to the change in its kinetic energy. In other words, when a force is applied to an object and causes it to move, the work done by that force is equal to the change in the object's speed.

2. How is the work energy principle related to power?

The work energy principle and power are closely related because power is the rate at which work is done. In other words, power is the amount of work done per unit of time. This means that the more power an object has, the faster it can do work.

3. What is the unit of measurement for work and energy?

The unit of measurement for work and energy is the joule (J). One joule is equal to the amount of work done when a force of one newton is applied over a distance of one meter. Energy can also be measured in other units such as calories, kilowatt-hours, and foot-pounds.

4. Can the work energy principle be applied to all types of energy?

Yes, the work energy principle can be applied to all types of energy, including mechanical, electrical, and thermal energy. This is because all forms of energy can be converted into work, and vice versa, according to the principle of conservation of energy.

5. How is the work energy principle used in real-world applications?

The work energy principle is used in many real-world applications, such as calculating the power output of engines, designing roller coasters, and understanding the motion of objects in sports. It is also used in fields such as engineering, physics, and mechanics to analyze and solve problems involving work, energy, and power.

• General Math
Replies
16
Views
1K
• General Math
Replies
1
Views
864
• General Math
Replies
1
Views
563
• General Math
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
8
Views
3K
• Classical Physics
Replies
35
Views
2K
• Introductory Physics Homework Help
Replies
26
Views
3K
• Introductory Physics Homework Help
Replies
3
Views
2K
• Introductory Physics Homework Help
Replies
6
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
3K