Work energy principle and power

In summary: Yes, that is the correct way of calculating. You have correctly applied the conservation of energy principle to solve for the depth of the pond. Great job!
  • #1
Shah 72
MHB
274
0
A golf ball of mass 45.9g is hit from a tee with speed 50 m/ s. The ball lands in a pond 5m lower than the tee. When the ball lands in the pond it has traveled a curved path of length 160m. The resistance acting on the ball has magnitude 0.3N
a) Find the speed of the ball just before it hits the water
I got the ans 22.5m/ s

The water immediately absorbs 8J of energy from the ball. The ball then sinks vertically downwards to reach the bottom of the pond. The resistance acting on the ball had magnitude 3N and the ball just comes to rest as it reaches the bottom of the pond.
b) find the depth of the pond
I get the ans as 1. 22m . The ans in the textbook is 1.44m
Pls help
 
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  • #2
Please start showing the work you've done. We can't help you much if you don't show us what you are trying to do! If there is a mistake in your work we can correct it.

-Dan
 
  • #3
topsquark said:
Please start showing the work you've done. We can't help you much if you don't show us what you are trying to do! If there is a mistake in your work we can correct it.

-Dan
Sure.
So for the (a)
m=0.0459 kg, u= 50 m/ s, h =5m, s=160m, resistance = 0.3N
Increase in KE= 0.02295v^2- 57.375J
Increase in GPE= -2.295J
Work done against resistance = -48J
So v= 22.5m/ s
For (b)
As soon as the ball falls in the pool, 8J of energy is lost by the ball.
Resistance on the ball= 3N
Work done against resistance = - 3S
Energy of the ball before entering the pond= 11.63J
11.63-8= 3s
S=1.22m
 
  • #4
for (b), the resistance of 3N on the ball as it sinks is not the net force
 
  • #5
skeeter said:
for (b), the resistance of 3N on the ball as it sinks is not the net force
Can you pls show how to calculate. Iam not able to understand
 
  • #6
Shah 72 said:
Can you pls show how to calculate. Iam not able to understand

what are the forces acting on the ball as it sinks?
 
  • #7
skeeter said:
what are the forces acting on the ball as it sinks?
The forces acting on the ball is the final KE, final GPE and the resistance

So 11.62- 0.459h-8= -3h
 
  • #8
skeeter said:
what are the forces acting on the ball as it sinks?
Iam not sure if it's right
 
  • #9
kinetic energy is not a force …

48DC6887-B8C9-4EEA-98F7-1C2E59B0F5CB.jpeg
 
  • #10
skeeter said:
kinetic energy is not a force …

View attachment 11247
So contact force = 459N along with resistance of 0.3N before it enters water and 3N as it reaches water
 
  • #11
skeeter said:
kinetic energy is not a force …

View attachment 11247
Iam still very confused on how to calculate. Pls help
 
  • #12
from the very beginning ...

A golf ball of mass 45.9g is hit from a tee with speed 50 m/ s. The ball lands in a pond 5m lower than the tee. When the ball lands in the pond it has traveled a curved path of length 160m. The resistance acting on the ball has magnitude 0.3N
a) Find the speed of the ball just before it hits the water

let the zero for GPE be the water level

$KE_0 + mgh_0 - R_1 \cdot s_1 = KE_f$

where the magnitude of $R_1 = 0.3 \, N$, $s_1 = 160 \, m$

The water immediately absorbs 8J of energy from the ball. The ball then sinks vertically downwards to reach the bottom of the pond. The resistance acting on the ball had magnitude 3N and the ball just comes to rest as it reaches the bottom of the pond.
b) find the depth of the pond

$KE_f - 8 - (R_2 - mg) \cdot s_2 = 0$

where the magnitude of $R_2 = 3 \, N$ and and the weight of the ball, $mg$, is not 459N
 
  • #13
Shah 72 said:
Iam still very confused on how to calculate. Pls help
Please start telling us what you are confused about! We know you are confused... If you are mistaking kinetic energy for a force then you have a long way to go to solve one of these. But we aren't going to simply spoon feed you and guess where you need the help. Please start being more interactive. It will help you, even in the short run.

-Dan
 
  • #14
skeeter said:
from the very beginning ...
let the zero for GPE be the water level

$KE_0 + mgh_0 - R_1 \cdot s_1 = KE_f$

where the magnitude of $R_1 = 0.3 \, N$, $s_1 = 160 \, m$
$KE_f - 8 - (R_2 - mg) \cdot s_2 = 0$

where the magnitude of $R_2 = 3 \, N$ and and the weight of the ball, $mg$, is not 459N
When it hits the water,
Increase in KE = 1/2 mv^2-1/2mu^2= 11.62J
Increase in GPE= mg(h2-h1)=0.0459×10(0- h)= -0.459h
Work done against resistance = -3×h
Energy lost from the ball= 8J
11.62 - 0.459h -8 = -3h
h= 1.42m
Is this the correct way of calculating?
 
  • #15
topsquark said:
Please start telling us what you are confused about! We know you are confused... If you are mistaking kinetic energy for a force then you have a long way to go to solve one of these. But we aren't going to simply spoon feed you and guess where you need the help. Please start being more interactive. It will help you, even in the short run.

-Dan
I am confused about how to calculate once it hits the water. Iam not getting how to calculate the mechanical energy and the work done against resistance.
When it hits the water
m= 0.0459kg, u= 22.5m/s , v=0m/s, energy lost from the ball and absorbed by the water= 8J, resistance = 3N
 
  • #16
Shah 72 said:
I am confused about how to calculate once it hits the water. Iam not getting how to calculate the mechanical energy and the work done against resistance.
When it hits the water
m= 0.0459kg, u= 22.5m/s , v=0m/s, energy lost from the ball and absorbed by the water= 8J, resistance = 3N
Part b) Using romsek's notation in post #12:

It's the Work-Energy theorem.
\(\displaystyle W_{net} = \Delta KE\)

Step 1: The work done by gravity as it sinks: \(\displaystyle W_g = mgh - mgh_0 = -mg s_2\)

Step 2: The work done by the water: \(\displaystyle W_w = F s_2 = -R s_2\)

Step 3: 8 J of energy is absorbed by the water so there is an extra work done of 8 J.

Step 4: Using the final kinetic energy from part a) as the initial energy of part b):

So
\(\displaystyle -mg s_2 - R s_2 + 8 = - KE_f\)
which is romsek's equation.

If you are still confused, what part are you still confused about?

-Dan
 
  • #17
topsquark said:
Part b) Using romsek's notation in post #12:

It's the Work-Energy theorem.
\(\displaystyle W_{net} = \Delta KE\)

Step 1: The work done by gravity as it sinks: \(\displaystyle W_g = mgh - mgh_0 = -mg s_2\)

Step 2: The work done by the water: \(\displaystyle W_w = F s_2 = -R s_2\)

Step 3: 8 J of energy is absorbed by the water so there is an extra work done of 8 J.

Step 4: Using the final kinetic energy from part a) as the initial energy of part b):

So
\(\displaystyle -mg s_2 - R s_2 + 8 = - KE_f\)
which is romsek's equation.

If you are still confused, what part are you still confused about?

-Dan
The ans is 1.44m.
Work done by gravity= -0.459h2
Work
topsquark said:
Part b) Using romsek's notation in post #12:

It's the Work-Energy theorem.
\(\displaystyle W_{net} = \Delta KE\)

Step 1: The work done by gravity as it sinks: \(\displaystyle W_g = mgh - mgh_0 = -mg s_2\)

Step 2: The work done by the water: \(\displaystyle W_w = F s_2 = -R s_2\)

Step 3: 8 J of energy is absorbed by the water so there is an extra work done of 8 J.

Step 4: Using the final kinetic energy from part a) as the initial energy of part b):

So
\(\displaystyle -mg s_2 - R s_2 + 8 = - KE_f\)
which is romsek's equation.

If you are still confused, what part are you still confused about?

-Dan
If I substitute the values in the equation you have given, it will be
-0.459h2-0.459h2 +8= -11.62
I get h2=21.4m
The ans in the textbook is 1.44m
 

Related to Work energy principle and power

1. 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 the net work done on an object will result in a change in its speed or direction of motion.

2. How is work calculated using the work-energy principle?

Work is calculated by multiplying the force applied to an object by the distance it 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.

3. What is the relationship between work and energy?

Work and energy are closely related concepts. Work is the transfer of energy from one object to another, or the change in energy of an object. Energy is the ability to do work, and can exist in various forms such as kinetic, potential, or thermal energy.

4. How does power relate to the work-energy principle?

Power is the rate at which work is done, or the amount of work done in a given amount of time. It is calculated by dividing work by time, or P = W/t. The work-energy principle can also be used to calculate power by dividing the change in an object's kinetic energy by the time it takes to change.

5. How is the work-energy principle applied in real-world situations?

The work-energy principle is applied in various fields such as physics, engineering, and sports. For example, it is used to calculate the amount of force needed to push a car up a hill, or the amount of energy required to launch a rocket into space. In sports, it is used to analyze the performance of athletes and their use of energy during activities such as running or throwing.

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