Speed of a box at the bottom of a incline.

In summary: For instance, if you have someone jumping off a cliff and you want to calculate their velocity at various points in their jump, you would need to use a kinetic energy equation like the one you found.
  • #1
Paulbird20
53
0
Ok so I am having trouble calculating the speed of a box as it travels down a incline.

I was Given 22 degree angle and the coefficient of friction (kinetic) is .22. Height of 9.10m (from rest at the top).
I calculated acceleration while it travels down the slop and got 1.6721.
I next need to calculate the speed when it reaches the bottom of the incline.

I thought i would simply need to use 2*G*H ^(1/2) . It gives me 13.355 and it is wrong.
Any help would be greatly appreciated or any equations i can use.

Thanks from a new user.
Paul.
 
Physics news on Phys.org
  • #2
Why not use the approximation of gravitational potential energy for objects close to Earth?

U[g]=mgh

But that doesn't just equal the kinetic energy at the bottom. You have to take into account the work done by nonconservative forces, namely, in this case, friction.

I don't know if that was any help or not.
 
  • #3
I don't know how to calculate the mass to be able to use that equation. Only equations i have in my notes are
2*G*H^(1/2) = V1
and
calculations of sum of all forces. I have a free body diagram drawn also i just don't know where to go from here. I was almost sure 13.355 was gunan be the speed but it is wrong.
 
  • #4
That equation is true if the block is frictionless but it is not frictionless in this case. Go back and re-read your notes or the book to better understand the U[g]=mgh gravitational potential energy equation.

The equation that you have comes from conservation of energy:

(mv^2)/2=mgh; rearrange that
v=(2gh)^1/2

But, as I said before, you cannot apply that equation here because the surface is not frictionless. Some energy is lost to friction.

Here's a hint: W[friction] =F[friction]*distance

Subtracting this value from your initial potential energy will give you the final kinetic energy at the bottom of the block.
 
  • #5
Ok so from what i have gathered reviewing my notes.

Fnormal= m*g*cos (angle)
and
FF(friction force) = coef K * Fn

I was trying to use these two equations to determine FF but i can't because of the mass in the first one. Is there another equation i can use to get FF?
 
  • #6
AH HA! i found my equation your hint helped thank you.

I took 2 * A* x1-x0 ^(1/2) and it gave me my final velocity Thank you.
 
  • #7
Ah, I see. You used a kinematics equation. Just remember, it may have worked in this case but those five major equations only work when the acceleration is a constant value.
 

1. What is the formula for calculating the speed of a box at the bottom of an incline?

The formula for calculating the speed of a box at the bottom of an incline is v = √(2gh), where v is the speed, g is the acceleration due to gravity (9.8 m/s²), and h is the height of the incline.

2. How does the mass of the box affect its speed at the bottom of an incline?

The mass of the box does not affect its speed at the bottom of an incline. The speed is solely determined by the height of the incline and the acceleration due to gravity.

3. Is the speed of the box at the bottom of an incline affected by the angle of the incline?

Yes, the speed of the box at the bottom of an incline is affected by the angle of the incline. The steeper the incline, the faster the box will accelerate and the greater its speed will be at the bottom.

4. Can the speed of the box at the bottom of an incline be greater than the speed of a falling object?

No, the speed of the box at the bottom of an incline cannot be greater than the speed of a falling object. This is because the box is still affected by the force of gravity as it moves down the incline, whereas a falling object is only affected by gravity.

5. How does friction impact the speed of a box at the bottom of an incline?

Friction can slow down the speed of a box at the bottom of an incline. This is because friction acts in the opposite direction of the box's motion, causing it to lose some of its kinetic energy and therefore decreasing its speed.

Similar threads

  • Introductory Physics Homework Help
Replies
15
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
25
Views
453
  • Introductory Physics Homework Help
Replies
11
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
814
  • Introductory Physics Homework Help
Replies
21
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
362
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
1K
Back
Top