Finding time and distance with NLM.

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A puck with a mass of 30 g slides on rough ice, experiencing a frictional force of 0.20 N while initially moving at 10 km/h (2.7 m/s). The calculations show that it takes 0.40 seconds for the puck to stop. However, the initial calculation for distance was incorrect; the average velocity must be considered since the puck decelerates from 2.7 m/s to 0 m/s. The correct distance traveled on the ice patch is 2.16 meters. The discussion emphasizes the importance of using average velocity in such scenarios.
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Homework Statement



A puck of mass 30 g slides across rough ice, experiencing a frictional force of 0.20N. If it was moving at 10km/h when it hit the ice patch,
(a) how much time did it take to stop?
(b) how long was the ice patch?

Given:
m=30g (0.03kg)
Ff=0.20N
V=10km/h (2.7m/s)[/B]

Homework Equations


Fnet=ma
a=v/t
v=d/t[/B]

The Attempt at a Solution



Fnet = ma
a = Fnet/m
a = 0.20N / 0.03kg
a = 6.6m/s2[/B]

a = v/t
t = v/a
t = 2.7m/s / 6.6m/s2
t = 0.40s

v = d/t
d = vt
d = (2.7m/s)(0.40s)
d = 1.08m

So therefore, it took 0.40s for the puck to stop. The ice patch was 1.08m.
 
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Sace Ver said:

Homework Statement



A puck of mass 30 g slides across rough ice, experiencing a frictional force of 0.20N. If it was moving at 10km/h when it hit the ice patch,
(a) how much time did it take to stop?
(b) how long was the ice patch?

Given:
m=30g (0.03kg)
Ff=0.20N
V=10km/h (2.7m/s)[/B]

Homework Equations


Fnet=ma
a=v/t
v=d/t[/B]

The Attempt at a Solution



Fnet = ma
a = Fnet/m
a = 0.20N / 0.03kg
a = 6.6m/s2[/B]

a = v/t
t = v/a
t = 2.7m/s / 6.6m/s2
t = 0.40s

v = d/t
d = vt
d = (2.7m/s)(0.40s)
d = 1.08m

So therefore, it took 0.40s for the puck to stop. The ice patch was 1.08m.
Can someone check it over PLS and thank you!
 
Sace Ver said:

Homework Statement



A puck of mass 30 g slides across rough ice, experiencing a frictional force of 0.20N. If it was moving at 10km/h when it hit the ice patch,
(a) how much time did it take to stop?
(b) how long was the ice patch?

Given:
m=30g (0.03kg)
Ff=0.20N
V=10km/h (2.7m/s)[/B]

Homework Equations


Fnet=ma
a=v/t
v=d/t[/B]

The Attempt at a Solution



Fnet = ma
a = Fnet/m
a = 0.20N / 0.03kg
a = 6.6m/s2[/B]

a = v/t
t = v/a
t = 2.7m/s / 6.6m/s2
t = 0.40s

v = d/t
d = vt
d = (2.7m/s)(0.40s)
d = 1.08m

So therefore, it took 0.40s for the puck to stop. The ice patch was 1.08m.
The time is correct. The distance is not correct.

The velocity is not constant, so the average velocity is not 2.7 m/s.
 
SammyS said:
The time is correct. The distance is not correct.

The velocity is not constant, so the average velocity is not 2.7 m/s.
So would 2.7m/s be the initial velocity and then 0 would be my final velocity?
 
Sace Ver said:
So would 2.7m/s be the initial velocity and then 0 would be my final velocity?
The puck comes to a stop, so yes.
 
SammyS said:
The puck comes to a stop, so yes.
so it'd be d = 2.16?
 
Sace Ver said:
so it'd be d = 2.16?
Yes.
 
SammyS said:
Yes.
THANK U!
 
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