Help velocity and mass given

In summary, the problem involves a 1.00 kg disk sliding on a horizontal steel plate with an initial velocity of 5.00 m/s. The coefficient of friction is given as 0.04 and the goal is to find the time it takes for the disk to stop and the distance it travels. Using the equation a = (V2 - V1)/T, the acceleration is determined to be -0.3924 m/s^2 and the time is found to be 12.74 seconds. Using the equation d = v_o t + 1/2 a t^2, the distance is calculated to be 31.86 meters.
  • #1
Vick007
9
0

Homework Statement



With an initial speed of 5.00 m/s a 1.00 kg disk is slid on a horizontal steel plate. How long will it take for the disk to stop, and how far does it travel? ( do not neglect friction )

Homework Equations


I don't know how to use the mass in any equations. help.


The Attempt at a Solution


something with xo= vo + 1/2at^2 ?? how is the mass used?
 
Physics news on Phys.org
  • #2
Vick007 said:

Homework Statement



With an initial speed of 5.00 m/s a 1.00 kg disk is slid on a horizontal steel plate. How long will it take for the disk to stop, and how far does it travel? ( do not neglect friction )

Homework Equations


I don't know how to use the mass in any equations. help.


The Attempt at a Solution


something with xo= vo + 1/2at^2 ?? how is the mass used?

The question appears to be incomplete. You need to know the coefficient of dynamic friction between the disk and steel plate in order to proceed. You'll use the coefficient of friction and the mass to determine the force that retards the motion, hence the acceleration.
 
  • #3
ok so I checked and yeah the coefficient is given as 0.04 what do I do now? thanks
 
  • #4
Vick007 said:
ok so I checked and yeah the coefficient is given as 0.04 what do I do now? thanks

What is the relationship between the coefficient of friction and the frictional force?
 
  • #5
weight X coefficient of friction = friction force?
how does that tie in with the velocity given?
 
  • #6
Vick007 said:
weight X coefficient of friction = friction force?
Yes.
how does that tie in with the velocity given?

Draw the free body diagram for the disk. What forces are acting? What is the acceleration of the disk (hint: Newton's second law).
 
  • #7
Ff= force of friction
U= coefficient of friction
Fn= force normal
m=mass
g=acceleration due to gravity =9.81
a= acceleration
d=distance

Ff=UxFn
which is the same as
m x a= U x m x -g
divide out the m and you get
a=U x -g
and so
a=.04 x -9.81
a= -.3924
so is it slowing down by .3924 m/s^2?
how do I link the velocity, time, and acceleration now?
 
  • #8
Vick007 said:
Ff= force of friction
U= coefficient of friction
Fn= force normal
m=mass
g=acceleration due to gravity =9.81
a= acceleration
d=distance

Ff=UxFn
which is the same as
m x a= U x m x -g
divide out the m and you get
a=U x -g
and so
a=.04 x -9.81
a= -.3924
so is it slowing down by .3924 m/s^2?
how do I link the velocity, time, and acceleration now?

What's the relationship between acceleration and velocity? How about if there's an initial velocity?
 
  • #9
the initial velocity is 5.00 m/s and since the acceleration is (I think) -.3924 and I'm looking for the time would this equation work?
a = (V2 - V1)/T
where
(V2 - V1) = change in velocity
so it would say
-.3924 = (V2 - 5.00)/T
but what is V2? or how do I find it?
 
  • #10
Vick007 said:
the initial velocity is 5.00 m/s and since the acceleration is (I think) -.3924 and I'm looking for the time would this equation work?
a = (V2 - V1)/T
That will work. It's a rearrangement of the usual equation, v2 = v1 + a*t .
where
(V2 - V1) = change in velocity
so it would say
-.3924 = (V2 - 5.00)/T
but what is V2? or how do I find it?

What is the speed of the puck when it stops?
 
  • #11
so V2 = 0
and now I just have to find T
so according to your equation it would look like this:
v2 = v1 + a*t
and if I fill it in I get
0 = 5.00 + (-.3924) * t
-5 = -.3924 * t
(-5/-.3924) = t
t = 12.42 sec (is that correct?)
thanks a lot for helping, but there's a second part to the problem
How long will it take for the disk to stop, and how far does it travel? ( do not neglect friction )
how do I begin the second part now that I have the time? Is there some equation?
 
  • #12
Vick007 said:
so V2 = 0
and now I just have to find T
so according to your equation it would look like this:
v2 = v1 + a*t
and if I fill it in I get
0 = 5.00 + (-.3924) * t
-5 = -.3924 * t
(-5/-.3924) = t
t = 12.42 sec (is that correct?)
Probably a typo, t = 12.74 should be your value.
thanks a lot for helping, but there's a second part to the problem
How long will it take for the disk to stop, and how far does it travel? ( do not neglect friction )
how do I begin the second part now that I have the time? Is there some equation?

I'm rather surprised that you don't seem to have had any introduction to the basic kinematic formulas. Yes there is an equation (there's always an equation :smile ). In this case you want one that relates distance traveled to initial velocity, acceleration, and time. Any ideas as to which of the standard kinematic equations might fit?
 
  • #13
would it be one like this?
X - Xo = .5(Vo + V)t
which could be simplified to
X = .5(5+V)(12.74)
how do I find the velocity though?
 
  • #14
Vick007 said:
would it be one like this?
X - Xo = .5(Vo + V)t
which could be simplified to
X = .5(5+V)(12.74)
how do I find the velocity though?

That equation will work. It incorporates the initial velocity (Vo) and the final velocity (V = 0) in order to take care of the acceleration. Remember that you previously wrote: a = (V2 - V1)/T.

So plug in your V=0 and find distance X.

Another equation that you should become familiar with is:

[itex]d = v_o t + \frac{1}{2} a t^2 [/itex]

See the Physics Formulary
 
  • #15
ok so I got like
X = 31.86 meters?
is that correct?
 
  • #16
Vick007 said:
ok so I got like
X = 31.86 meters?
is that correct?

Yes, that is correct.
 

1. What is the equation for calculating velocity?

The equation for calculating velocity is velocity = distance / time. This means that velocity is equal to the distance an object travels divided by the time it takes to travel that distance.

2. How is mass related to velocity?

Mass and velocity are directly related, meaning that as mass increases, velocity decreases and vice versa. This is because mass is a measure of an object's inertia, or resistance to changes in motion, and a larger mass requires more force to accelerate and reach a certain velocity.

3. Can an object's velocity change without changing its mass?

Yes, an object's velocity can change without changing its mass. This is because velocity is affected by factors such as force, distance, and time, not just mass. For example, an object can change velocity by altering the amount of force acting on it or by changing the distance it travels in a certain amount of time.

4. How does the mass of an object affect its velocity in free fall?

In free fall, all objects, regardless of their mass, accelerate at the same rate due to gravity. This means that the mass of an object does not affect its velocity in free fall. However, objects with greater mass will experience a greater force of gravity, which can impact their velocity upon impact with the ground.

5. What units are used to measure velocity and mass?

Velocity is typically measured in units of distance per time, such as meters per second (m/s) or kilometers per hour (km/h). Mass is measured in units of mass, such as kilograms (kg) or grams (g). In scientific calculations, it is important to use consistent units for both velocity and mass to ensure accurate results.

Similar threads

  • Introductory Physics Homework Help
Replies
34
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
763
  • Introductory Physics Homework Help
Replies
33
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
763
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
3K
  • Introductory Physics Homework Help
Replies
18
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
944
Back
Top