Acceleration of a Block on an Inclined Plane

In summary, a block of mass 5.5 kg is being pulled up a 22° incline with a force of 38 N. If the incline is frictionless, the acceleration of the block is 3.24 m/s2. If the coefficient of kinetic friction is 0.12, the acceleration is 2.159638636 m/s2. The normal force is perpendicular to the incline and can be found by analyzing force components. However, the weight of the block, which is vertical, will also have a normal component.
  • #1
riseofphoenix
295
2
7. A block of mass m = 5.5 kg is pulled up a θ = 22° incline as in the figure with a force of magnitude F = 38 N.

4-p-030.gif


(a) Find the acceleration of the block if the incline is frictionless

So I got this one right, and here is what I did...

From the free body diagram I drew:

Fnet = (38 N) + (-mg sin 22)
ma = 38 - mg sin 22
a = (38 - mg sin 22)/m
a = [ 38 - (5.5)(9.81)(sin 22) ] / (5.5)
a = (38 - 20.211) / 5.5
a = 3.24 m/s2

(b) Find the acceleration of the block if the coefficient of kinetic friction between the block and incline is 0.12.

So...I know that

fk = μkn
fk = 0.12(Fappliedsin θ - mg)

QUESTION 1: So this is where I'm a little confused...Normal force should be UPWARDS, right? then would that mean that n = Fappliedsin θ + (- mg) after rotating my free body diagram so that I can see the axis better? Because I almost got the answer...

fk = 0.12[38 sin 22 - (5.5)(9.81)]
fk = -4.767 (the negative number makes sense since kinetic friction is OPPOSING the motion of the block which is moving up the incline.

So then I add it to the equation for acceleration at the top

ma = (38 N) + (-mg sin 22) + (-4.767)
a = [ 38 - (5.5)(9.81)(sin 22) - 4.767 ] / (5.5)
a = 13.022 / 5.5
a = 2.36 m/s2

This answer is wrong...Webassign says the answer is actually a = 2.15 m/s2

QUESTION 2: Where did I go wrong? Was it the normal force equation part?
 
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  • #2


riseofphoenix said:
So...I know that

fk = μkn
Right.
fk = 0.12(Fappliedsin θ - mg)
Not right.
QUESTION 1: So this is where I'm a little confused...Normal force should be UPWARDS, right? then would that mean that n = Fappliedsin θ + (- mg) after rotating my free body diagram so that I can see the axis better? Because I almost got the answer...
The normal force is 'normal'--perpendicular--to the surface. Figure out the normal force by analyzing force components perpendicular to the incline. They must add to zero.
 
  • #3


Doc Al said:
Right.

Not right.

The normal force is 'normal'--perpendicular--to the surface. Figure out the normal force by analyzing force components perpendicular to the incline. They must add to zero.

No wait...
it would be

-mg + (-mg sin 22)?
 
  • #4


Doc Al said:
Right.

Not right.

The normal force is 'normal'--perpendicular--to the surface. Figure out the normal force by analyzing force components perpendicular to the incline. They must add to zero.

It's just F cos θ, which is 38 cos 22...

But I still don't get the right answer when I multiply 0.12 by 38 cos 22, and then add 4.22 to this equation...

ma = 38 - 20.11

a = (38 - 20.11 + 4.22) / 5.5 doesn't give me 2.15 m/s2

Something else is wrong
 
Last edited:
  • #5


riseofphoenix said:
It's just F cos θ, which is 38 cos 22...
The applied force F is parallel to the incline. It has no normal component.

The weight of the block, being vertical, will have a normal component.
 
  • #6


Doc Al said:
The applied force F is parallel to the incline. It has no normal component.

The weight of the block, being vertical, will have a normal component.

So it's -mg cos 22?
Which is basically -(5.5)(9.81)(cos 22)...

fk = 0.12[-(5.5)(9.81)(cos 22)]
fk = -6.003

So,

a = [ 38 - 20.211 + (-6.003) ] / 5.5
a = 2.159638636

:)

Thanks!
 
  • #7


riseofphoenix said:
So it's 5.5 cos 22?
Almost. What's the weight of the block?
 
  • #8


Doc Al said:
Almost. What's the weight of the block?

It's just 5.5 kg isn't it?
 
  • #9


riseofphoenix said:
It's just 5.5 kg isn't it?
No, that's the mass.
 
  • #10


Doc Al said:
No, that's the mass.

oh...Weight = mg
Weight = -(5.5)(9.81) = -53.955 N
 
  • #11


riseofphoenix said:
oh...Weight = mg
Weight = -(5.5)(9.81) = -53.955 N
Right. The weight = mg and acts downward.
 
  • #12


Doc Al said:
Right. The weight = mg and acts downward.

Ohh ok!
Thanks a lot!
 

What is the formula for calculating the acceleration of a block on an inclined plane?

The formula for calculating the acceleration of a block on an inclined plane is a = gsinθ, where a is the acceleration, g is the acceleration due to gravity (9.8 m/s^2), and θ is the angle of inclination.

Does the mass of the block affect its acceleration on an inclined plane?

Yes, the mass of the block does affect its acceleration on an inclined plane. The greater the mass of the block, the greater the force of gravity acting on it, resulting in a greater acceleration down the plane.

How does the angle of inclination affect the acceleration of the block?

The angle of inclination affects the acceleration of the block by changing the component of gravity acting down the plane. As the angle increases, the component of gravity acting down the plane also increases, resulting in a greater acceleration.

Can the acceleration of the block on an inclined plane be greater than the acceleration due to gravity?

Yes, the acceleration of the block on an inclined plane can be greater than the acceleration due to gravity. This occurs when the angle of inclination is greater than 45 degrees, resulting in a greater component of gravity acting down the plane.

Is there a difference in acceleration between a block sliding down an inclined plane and a block rolling down an inclined plane?

Yes, there is a difference in acceleration between a block sliding down an inclined plane and a block rolling down an inclined plane. The block rolling down the plane experiences both translational and rotational motion, resulting in a lower acceleration compared to a block sliding down the plane with only translational motion.

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