Finding time and distance with NLM.

In summary, the puck of mass 30 g took 0.40 seconds to stop while sliding at 10km/h on rough ice, experiencing a frictional force of 0.20N. The ice patch was approximately 2.16 meters long.
  • #1
Sace Ver
79
2

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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  • #2
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!
 
  • #3
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.
 
  • #4
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?
 
  • #5
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.
 
  • #6
SammyS said:
The puck comes to a stop, so yes.
so it'd be d = 2.16?
 
  • #7
Sace Ver said:
so it'd be d = 2.16?
Yes.
 
  • #8
SammyS said:
Yes.
THANK U!
 

Related to Finding time and distance with NLM.

1. How does Newton's Laws of Motion (NLM) help us find time and distance?

Newton's Laws of Motion provide the basic principles and equations for calculating time and distance in relation to an object's motion. The first law states that an object in motion will stay in motion unless acted upon by an external force. This means that if an object is moving at a constant speed, it will continue to do so until a force changes its motion. The second law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This allows us to calculate the acceleration, which is necessary for finding time and distance. The third law states that for every action, there is an equal and opposite reaction. This helps us understand the forces acting on an object, which can affect its time and distance traveled.

2. How do we calculate time using NLM?

To calculate time using NLM, we need to know the acceleration and initial velocity of an object. Using the equation t = v/a, where t is time, v is initial velocity, and a is acceleration, we can determine how long it will take for an object to travel a certain distance. This equation assumes that the acceleration is constant. If the acceleration is changing, we need to use calculus to find the time.

3. What is the relationship between distance and time in NLM?

In NLM, the distance an object travels is directly proportional to the square of the time it takes to travel that distance. This is represented by the equation d = 1/2 at^2, where d is distance, a is acceleration, and t is time. This relationship is important because it shows that the farther an object travels, the longer it will take, and the faster an object travels, the farther it will go in a given amount of time.

4. How can we use NLM to find the distance traveled by an object?

To find the distance traveled by an object using NLM, we need to know the initial velocity, acceleration, and time. Using the equation d = v*t + 1/2at^2, we can calculate the distance traveled by an object. This equation takes into account the initial velocity and the acceleration, which can affect the distance traveled. If the acceleration is constant, we can use the simpler equation d = v*t, where d is distance, v is initial velocity, and t is time.

5. Can NLM be used to find the time and distance for any type of motion?

Yes, NLM can be used to find the time and distance for any type of motion, as long as the acceleration is constant. This includes linear motion, circular motion, and even projectile motion. However, if the acceleration is not constant, we need to use more complex equations or calculus to find the time and distance. NLM provides the basic principles and equations for understanding motion and can be applied to a wide range of scenarios.

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