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

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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.
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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!
 
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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!
 

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