Dynamics Problem HELP - Free body diagram

AI Thread Summary
The discussion revolves around a physics problem involving the dynamics of a golf ball on a sloped green, specifically comparing uphill and downhill putts. The user is struggling with calculations and free body diagrams to determine which putt is more difficult, noting that uphill putts require more force. There are concerns about the accuracy of the calculations presented, particularly regarding the use of units and the interpretation of forces involved. The conversation highlights the need for clear definitions of variables and proper application of kinematic equations to find the initial velocities required for the ball to stop within 1.0m of the hole. Overall, the user seeks guidance on both the mathematical reasoning and the correct approach to the problem.
coglon
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IM in Phys.20 Got this question for an assignment. Tried it and got stuck. I really need help
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When golfing, uphill and downhill putts present different levels of difficulty. Suppose a green has a 5.0 degree slope and the force of friction will be 0.10N against the ball's motion. The ball has a mass of 46g and the length of putt will be 10.0m.

Which putt (uphill or downhill) presents a greater level of difficulty? Explain your reasoing using appropriate calculations

Draw Free body diagrams for both purrs. Find range of speed you may give the ball on each type of putt so that even if you miss the ball will stop within 1.0m of the hole.
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I know uphill would be more difficult because it requires more force but I am having a hard time proving this using Free body diagrams and mathematical reasoning.
THANKS FOR THE HELP,
Coglon
 
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Dynamics -- Still stuck.. need ideas

Fm=Fa+Fg
46^2=Fa+9.81^2
Fa=[squ]2019.7639
fa= 44.9m/s^2
F=Ma
=46g*44.9m/s^2
=2065N

Using Free body diagrams I believe I figured out that it would take...
43.8N downhill and 64.3N uphill
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Im not sure if the above calculation are at all close and I cannot show you my Free Body Diagram due to the attach file would be too large and would not be readable to fit in attachment... I am have no clue how to determine the range of speed used to get the putt within 1.0m and whether or not I should be using some kinematic equations.

Thanks for help,
Coglon
 


Hi coglon,

welcome to the forums.

This is an interresting problem...

Originally posted by coglon
Fm=Fa+Fg
46^2=Fa+9.81^2
Fa=[squ]2019.7639
fa= 44.9m/s^2
F=Ma
=46g*44.9m/s^2
=2065N

Uh, not exactly sure what you're trying to do here. What do Fm, Fa, and Fg represent?

You also need to check your units. A Newton is 1kg*1m/s2, not 1g*1m/s2

Using Free body diagrams I believe I figured out that it would take...
43.8N downhill and 64.3N uphill

You aren't going to be able to supply it with a force over the entire length of the path. The problem is assuming you bump it to give it an initial velocity (not caring the physics of the bump), and then applying friction and gravity forces to slow down/speed up the ball.

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Im not sure if the above calculation are at all close and I cannot show you my Free Body Diagram due to the attach file would be too large and would not be readable to fit in attachment...

What type of file is it?

I am have no clue how to determine the range of speed used to get the putt within 1.0m and whether or not I should be using some kinematic equations.

The way I see it, you'll need to do 4 different equations for two different cases. You will know the final velocity (0), and through the free body diagrams you can find the total force for the two cases. You'll be solving for initial velocities given a final position of 9m and 11m (one meter past the hole and one meter in front of the hole) for each case.
 
I will second enigma in his wondering if you really have looked carefully at what you are doing. In addition to writing Fm= Fa+Fg without telling us what Fm, Fa, Fg mean (I would guess that Fg is "force of gravity", perhaps Fm is the net force on the ball. Would Fa be the friction force?) you have replaced Fm with (46)2 and Fg with (9.81)2. The only "46" in the problem is mass of the ball: 46 grams and I would guess that "9.81" is the acceleration due to gravity. Neither (46g)2 nor (9.81 m/s2)2 is a force and since they don't have the same units, they certainly would not fit into that equation no matter what Fm, Fa, or Fg were supposed to be.
 
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