Calculate Force of Friction on Hockey Puck Slowing to 10m/s

  • Thread starter Thread starter Kennedy111
  • Start date Start date
  • Tags Tags
    Energy
Click For Summary
To calculate the force of friction on a hockey puck slowing from 25 m/s to 10 m/s over 7.5 m, the kinetic energy before and after the slowdown is determined. The initial kinetic energy is calculated as 2.125 J, while the final kinetic energy is 0.85 J, resulting in a work done of -1.325 J. The force of friction is then calculated using the work-energy principle, yielding a value of approximately -0.1767 N. A user expressed confusion about their calculations, but another participant pointed out an error in their approach. This discussion highlights the importance of accurately applying kinetic energy and work equations in physics problems.
Kennedy111
Messages
27
Reaction score
0
Energy Question! Please Help!

Homework Statement


A 170g hockey puck sliding at 25 m/s slows to a speed of 10 m/s in a distance of 7.5m. Determine the force of friction causing the puck to slow down.

m=170g
= 0.170kg
V1 = 25 m/s
V2= 10 m/s
d = 7.5 m/s

Homework Equations


Kinetic Energy = (1/2)(m)(v^2)
Work = Fd

The Attempt at a Solution



I'm really confused...

Kinetic Energy 1 = (1/2)(0.170kg)(25m/s)
= 2.125 J
Kinetic Energy 2 = (1/2)(0.170kg)(10m/s)
= 0.85 J

Kinetic Energy 2 - Kinetic Energy 1
0.85 J - 2.125 J = -1.325 J

Work = Fd
F = Work/d
= -1.325/7.5m
= -0.176666... which is wrong!

Please help!
 
Physics news on Phys.org


Kennedy111 said:

Homework Statement


A 170g hockey puck sliding at 25 m/s slows to a speed of 10 m/s in a distance of 7.5m. Determine the force of friction causing the puck to slow down.

m=170g
= 0.170kg
V1 = 25 m/s
V2= 10 m/s
d = 7.5 m/s

Homework Equations


Kinetic Energy = (1/2)(m)(v^2)
Work = Fd


The Attempt at a Solution



I'm really confused...

Kinetic Energy 1 = (1/2)(0.170kg)(25m/s)
= 2.125 J
Kinetic Energy 2 = (1/2)(0.170kg)(10m/s)
= 0.85 J

Kinetic Energy 2 - Kinetic Energy 1
0.85 J - 2.125 J = -1.325 J

Work = Fd
F = Work/d
= -1.325/7.5m
= -0.176666... which is wrong!

Please help!

Check you equation.
v2
 


Oh duh! That was dumb of me! Thank you so much!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 6 ·
Replies
6
Views
14K
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 8 ·
Replies
8
Views
4K
Replies
18
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
17
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K