How Do You Calculate Force and Acceleration on an Inclined Plane?

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SUMMARY

The discussion focuses on calculating the acceleration and force required to push a 60 kg block up a 26.5° inclined plane with a coefficient of kinetic friction of 0.17. The acceleration was determined using the equation Delta X = Vit + (1/2)at^2, yielding an acceleration of 0.069 m/s². The force of push was calculated using the net force equation, Fnet = ma, and the frictional force equation, Fkinetic = μ * N, leading to a final force of approximately 24.36 N after correctly determining the normal force acting on the block.

PREREQUISITES
  • Understanding of Newton's Second Law (Fnet = ma)
  • Knowledge of kinematic equations, specifically Delta X = Vit + (1/2)at²
  • Familiarity with frictional force calculations (Fkinetic = μ * N)
  • Ability to resolve forces into components on an inclined plane
NEXT STEPS
  • Study how to resolve gravitational forces into components on an inclined plane
  • Learn about the implications of different coefficients of friction on motion
  • Explore advanced kinematic equations for varying acceleration scenarios
  • Investigate the effects of angle and mass on the forces acting on objects on inclines
USEFUL FOR

Students in physics or engineering courses, educators teaching mechanics, and anyone interested in understanding dynamics on inclined planes.

swede5670
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Homework Statement


The 60 kg block starting at rest is pushed 7.8 m up a ramp at an angle θ = 26.5° to the horizontal in 15 s. If the coeficient of kinetic friction is 0.17:
What is the acceleration of block?
What is the force used to push the block up the ramp?

Homework Equations


Fnet = ma
Delta X = Vit + (1/2)at^2
Fg=ma
Fkinetic = μ * N

The Attempt at a Solution


I think I understand the first part,
7.8 = 0*15 + (1/2)(a)(15^2)
7.8 = 112.5a
.069=a

But the second part is what's giving me trouble
Do I need to make an Fnet = ma equation?
If I do I know that it would look like this
Force of Push - Force of friction = ma
Force of Push - Force of friction = (60)(.069)
Can I substitute in this equation? Fkinetic = μ * N for Fk?
Force of Push - (μ * N) = 60* .069
Fpush - (.17 * N) = 4.14

But what is my N force? I think that it would just be mass times acceleration of gravity and in that case:
Fpush - (.17 * (60 * 9.81)) = 4.14
Fpush - 100.06200 = 4.14
Fpush = 104.20200N

I'm not sure if this is right, I appreciate your help in advance.
 
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You did the first part right.

For the second part, you did everything right except your calculation of the normal force. Try drawing a force diagram of the block. On an inclined ramp, gravity doesn't act perpendicular to the plane of motion.
 
Do I just need to find the horizontal component of gravity?
 
Not the horizontal component, though you do need to break gravity into components. You need to find the component of gravity that is perpendicular to the ramp.
 
So then do I just do
Cosine (26.5) = A/H
and in this case adjacent is the force I am looking for and H is gravity
Is the gravity Fg? so I can substitute Fg=ma
60 * 9.81 = Fg

Then I have
Cosine (26.5) * (60 * 9.81) = A

Is this correct?
 
Alright well that's wrong and I'm not sure why
Cos(26.5) * H = A
Cos(26.5) * (60 x 9.81) = A
which ends up being 118.71N
When I plug that in I get
Fpush - (.17 x 118.97) = 4.14
Fpush = 24.36
And I this doesn't work, so I'm not sure what's going wrong
 

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