Solving the Physics of an Object's Slide

In summary: It is not correct, since you (1) used the mistyped -0.2 acceleration value to do the calculation, and (2) used a formula that I think is incorrect.What is the general formula you used, before you substituted values into it? I mean the formula that has vi2=______? My further assistance is contingent upon your providing this information.I used the formulas given in the problem. It is a little messy, but this is how it looks on paper:Δd = vf(t) + 1/2a(t)220 = 0.2(t) + 0.5(-0.4)(t)2I don't see that formula in
  • #1
barthayn
87
0

Homework Statement


An object is pushed along a rough horizontal surface and released. it slides for 10.0s before coming to rest and travels a distance of 0.2m during the last 1.0s of its slide. Assuming the acceleration to be uniform throughout
a) How fast was the object traveling upon release?
b) How fast was the object traveling when it reached the half way position in its slide?

vf = 0m/s
vi = ?
d1 = ?
d2 = 0.2m
a = ?
t = 10s

Homework Equations


a = Δv/Δt
Δd = vΔt = ((Vi+Vf)/2)Δt
vf = Δt-1/2a(Δt)2
Δd = vfΔt-1/2a(Δt)2

The Attempt at a Solution



a = -0.4/1
a = -0.4m/s2

Δd = (0.4(100))/2
Δd = 40/2
Δd = 20 m

vi2 = 0-(-0.2)(20.2)
vi2 = 4.04m2/s2
vi = 2.009m/s

Therefore the object was traveling at 2.009 m/s when first released.

For part b, I do not know where to begin. The acceleration is -0.4m/s2, as founded in the first part. But I do not know how to apply the information because I get a negative answer for the final velocity for the half way point.
 
Physics news on Phys.org
  • #2
barthayn said:

Homework Statement


An object is pushed along a rough horizontal surface and released. it slides for 10.0s before coming to rest and travels a distance of 0.2m during the last 1.0s of its slide. Assuming the acceleration to be uniform throughout
a) How fast was the object traveling upon release?
b) How fast was the object traveling when it reached the half way position in its slide?

vf = 0m/s
vi = ?
d1 = ?
d2 = 0.2m
a = ?
t = 10s

Homework Equations


a = Δv/Δt
Δd = vΔt = ((Vi+Vf)/2)Δt
vf = Δt-1/2a(Δt)2
Δd = vfΔt-1/2a(Δt)2

The Attempt at a Solution



a = -0.4/1
a = -0.4m/s2

Δd = (0.4(100))/2
Δd = 40/2
Δd = 20 m
Your notation is a little on the sloppy side, but you are basically okay up to here.

vi2 = 0-(-0.2)(20.2)
vi2 = 4.04m2/s2
vi = 2.009m/s
Where does this come from? It seems wrong to me.

Therefore the object was traveling at 2.009 m/s when first released.
I got a different answer. Can you explain the stuff beginning with "vi2=0-(-0.2)(20.2)" please?

For part b, I do not know where to begin. The acceleration is -0.4m/s2, as founded in the first part. But I do not know how to apply the information because I get a negative answer for the final velocity for the half way point.
A negative answer is a problem. Can you show your work, then I can help figure out what is wrong.
 
  • #3
Redbelly98 said:
Your notation is a little on the sloppy side, but you are basically okay up to here.


Where does this come from? It seems wrong to me.


I got a different answer. Can you explain the stuff beginning with "vi2=0-(-0.2)(20.2)" please?


A negative answer is a problem. Can you show your work, then I can help figure out what is wrong.

vi2=0-(-0.2)(20.2) is from the acceleration found, and the total distance traveled.

The final answer I got 2.009m/s. For b) I got 2.04m/s. To get that answer I did this:

Divided the total distance and time by two.

vf = (2d+a(t)2)/t
vf = (20.2-10)/5
vf = 10.2=5
vf = 2.04 m/s
 
  • #4
barthayn said:
vi2=0-(-0.2)(20.2) is from the acceleration found, and the total distance traveled.
The acceleration is -0.4 m/s2, but that value does not appear anywhere in your equation.
The distance traveled is 20.0 m, but you have 20.2 in your equation.
If you're using the equation I think you're using, check whether it has a factor of 1/2 or 2. OR you could use the simpler a=Δv/Δt that you listed as one of the Relevant Equations.:wink:

The final answer I got 2.009m/s. For b) I got 2.04m/s. To get that answer I did this:

Divided the total distance and time by two.
Actually, it is just the distance that is divided by two, not the time:
b) How fast was the object traveling when it reached the half way position in its slide?
"half way position" refers to the distance.
 
  • #5
Redbelly98 said:
The acceleration is -0.4 m/s2, but that value does not appear anywhere in your equation.
The distance traveled is 20.0 m, but you have 20.2 in your equation.
If you're using the equation I think you're using, check whether it has a factor of 1/2 or 2. OR you could use the simpler a=Δv/Δt that you listed as one of the Relevant Equations.:wink:


Actually, it is just the distance that is divided by two, not the time:

"half way position" refers to the distance.

You get the 0.2m plus from the question given. Also, in my notes, I got -0.4 m/s2. So I mistyped the number. The rest is correct.

How would I get the time to get it to reach the half way point. It states that the acceleration is uniform.
 
  • #6
barthayn said:
You get the 0.2m plus from the question given.
According to the problem statement, the object comes to rest in 10.0 s.
You correctly calculated that it moves 20 m in those 10 seconds. At that point it is at rest, so it cannot travel an additional 0.2 m. Instead, those 0.2 m comprise the final part of the 20 m total distance (and the final 1.0 s of the 10.0 s total).

Also, in my notes, I got -0.4 m/s2. So I mistyped the number. The rest is correct.
It is not correct, since you (1) used the mistyped -0.2 acceleration value to do the calculation, and (2) used a formula that I think is incorrect.

What is the general formula you used, before you substituted values into it? I mean the formula that has vi2=______? My further assistance is contingent upon your providing this information.

How would I get the time to get it to reach the half way point. It states that the acceleration is uniform.
You have acceleration and distance. Once part (a) is done correctly, you'll have initial velocity. Three quantities are sufficient to solve any 1-d uniform acceleration problem, you just need to find and use an equation that has a, Δd, and vi.
 
  • #7
Redbelly98 said:
According to the problem statement, the object comes to rest in 10.0 s.
You correctly calculated that it moves 20 m in those 10 seconds. At that point it is at rest, so it cannot travel an additional 0.2 m. Instead, those 0.2 m comprise the final part of the 20 m total distance (and the final 1.0 s of the 10.0 s total).It is not correct, since you (1) used the mistyped -0.2 acceleration value to do the calculation, and (2) used a formula that I think is incorrect.

What is the general formula you used, before you substituted values into it? I mean the formula that has vi2=______? My further assistance is contingent upon your providing this information.You have acceleration and distance. Once part (a) is done correctly, you'll have initial velocity. Three quantities are sufficient to solve any 1-d uniform acceleration problem, you just need to find and use an equation that has a, Δd, and vi.

The new first velocity I got was 4 m/s. Is this correct? For part b I got 2 m/s as well
 
  • #8
I agree with the 4 m/s. But it would be 2 m/s at half the total time, not half of the distance.

p.s. logging off for the night, will be back tomorrow.
 
  • #9
Redbelly98 said:
I agree with the 4 m/s. But it would be 2 m/s at half the total time, not half of the distance.

p.s. logging off for the night, will be back tomorrow.

How would I find out the time from this then? I am completely confused because the negative acceleration is uniform, meaning it is constant?
 
  • #10
Yes, the acceleration is constant.

You have Δd, a, and vi. Look through your list of kinematic equations for an equation that has those three terms in it.
 
  • #11
Redbelly98 said:
Yes, the acceleration is constant.

You have Δd, a, and vi. Look through your list of kinematic equations for an equation that has those three terms in it.

How would I find the time it took to reach the half way distance then? I do not understand how you can get it mathematically. Wouldn't it just be total time divided by two because it is consent negative acceleration.
 
  • #12
Not, it would have to be constant velocity for it to go half the total distance in half the total time. Hopefully that is common sense. This is not a constant velocity situation, since acceleration is not zero.

But -- guess what? -- you do not even need to know what that time is anyway. Use the kinematic equation that has Δd, a, vi (all of which you know) and vf (which you are trying to find).

Do you, or do you not, have the kinematic equations in your textbook or class notes?
 
  • #13
The final answer I got was 2.8 m/s. Is this correct?
 
  • #14
Yes, looks good to me :smile:
 

1. What factors affect an object's slide?

The factors that affect an object's slide include the object's mass, the surface it is sliding on, the force applied to the object, and the angle of the surface.

2. How does friction play a role in an object's slide?

Friction is the force that opposes the motion of an object, and it plays a significant role in an object's slide. The type of surface the object is sliding on and the force applied to the object can both affect the amount of friction present and ultimately impact the object's slide.

3. What is the difference between kinetic and static friction?

Kinetic friction is the force that opposes the motion of an object as it is already in motion. Static friction, on the other hand, is the force that must be overcome to set an object in motion. In other words, kinetic friction is present while the object is moving, while static friction is present when the object is at rest.

4. How can we calculate an object's acceleration during a slide?

The acceleration of an object during a slide can be calculated using Newton's second law of motion, which states that the net force on an object is equal to its mass multiplied by its acceleration (F=ma). By measuring the force applied to the object and its mass, we can calculate the acceleration during the slide.

5. How can we predict the distance an object will slide?

To predict the distance an object will slide, we can use the equations of motion, specifically the equation d=vi*t+1/2*a*t^2, where d is the distance, vi is the initial velocity, t is the time, and a is the acceleration. By plugging in the known values, we can calculate the distance the object will slide in a given amount of time.

Similar threads

  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
3K
  • Introductory Physics Homework Help
Replies
10
Views
4K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
9K
  • Introductory Physics Homework Help
Replies
12
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
3K
  • Introductory Physics Homework Help
Replies
27
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
2K
Back
Top