Compute acceleration on an inclined plane with friction

In summary, the problem involves a 24 kg object placed on an incline with a height of 500m and a length of 375m. The coefficient of kinetic friction is 0.600. The acceleration of the object is found to be 8.33646. To find the speed at the bottom of the incline, the equations for kinetic energy and potential energy are used and the final speed is found to be sqrt(2gh). The friction is not considered in this calculation.
  • #1
Jim Bob
4
0

Homework Statement


A 24 kg object is placed at the top of an incline with height 500m and length 375m. The coefficient of kinetic friction is 0.600
1) Find the acceleration of the object.
2) Find the speed by the time it has reached the bottom of the inclined plane.

Homework Equations


F_g = Force of gravity on the object
F_N = Normal force
F_// = Parallel force (force applied to the object parallel to the inclined plane and towards the bottom of the inclined plane)
F_p = Perpendicular force (force the object exerts on the inclined plane perpendicularly to it)
F_k = force of kinetic friction

The Attempt at a Solution


See the attachment, the answer key says acceleration=8.33646 but I got a different number
 

Attachments

  • PHYSICS MISTAKE.pdf
    37 KB · Views: 254
Physics news on Phys.org
  • #2
Jim Bob said:
height 500m and length 375m.
Length normally would mean the hypotenuse, but clearly that is not possible. I see you have taken the 375m as a horizontal extent, which is a reasonable interpretation.
This makes it a 3-4-5 triangle.
Jim Bob said:
answer key says acceleration=8.33646
That seems to be 85% of g. It would not be that high even without friction.
I agree with your answer, but you can get there much faster. The mass is irrelevant, and it is almost always unnecessary to find the angle.
You have (4/5)g-0.6(3/5)g=(11/25)g.
 
  • #3
What about the second problem? How would I do that?
 
  • #4
Jim Bob said:
What about the second problem? How would I do that?
What equations do you know that apply to motion under constant acceleration?
 
  • #5
Oh I got it now!, KE=mv^2/2 and PE = mgh so since all PE is transferred to KE by the end, mv^2/2=mgh and solving for v gives v=sqrt(2gh).
 
  • #6
Jim Bob said:
all PE is transferred to KE
What about the friction?
 

Related to Compute acceleration on an inclined plane with friction

1. What is the formula for calculating acceleration on an inclined plane with friction?

The formula for calculating acceleration on an inclined plane with friction is a = g(sinθ - μcosθ), where a is the acceleration, g is the gravitational constant (9.8 m/s^2), θ is the angle of incline, and μ is the coefficient of friction between the object and the surface.

2. How does friction affect the acceleration on an inclined plane?

Friction acts in the opposite direction of motion, so it can reduce the acceleration of an object on an inclined plane. The coefficient of friction determines the exact amount of acceleration reduction.

3. Can the acceleration on an inclined plane with friction be negative?

Yes, the acceleration on an inclined plane with friction can be negative if the angle of incline and coefficient of friction are such that the object is moving uphill. Negative acceleration indicates that the object is slowing down.

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

Kinetic friction is the force that opposes motion between two surfaces that are in contact and moving relative to each other. Static friction, on the other hand, is the force that must be overcome to set an object in motion when it is at rest on a surface. In the case of an inclined plane, kinetic friction would be in effect once the object starts moving, while static friction would apply if the object is not yet in motion.

5. How does the angle of incline affect the acceleration on an inclined plane with friction?

The greater the angle of incline, the greater the acceleration will be for an object on an inclined plane with friction. This is because a steeper incline creates a greater component of the gravitational force acting parallel to the surface, which results in a larger acceleration.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
1K
Replies
10
Views
520
  • Introductory Physics Homework Help
Replies
2
Views
787
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
11
Views
2K
  • Introductory Physics Homework Help
Replies
19
Views
2K
  • Introductory Physics Homework Help
Replies
33
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
409
Back
Top