Solved: Help with a block sliding down an Inclined platform

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In summary, the problem involves analyzing the motion of a block sliding down an inclined plane, considering factors such as gravitational force, friction, and the angle of inclination. The solution typically includes applying Newton's second law to determine the block's acceleration and final velocity, while also calculating the work done by friction and the potential energy converted to kinetic energy as the block descends. Understanding these concepts helps in predicting the block's behavior on the incline.
  • #1
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Homework Statement
A 5 kg block slides down a 30° incline with an acceleration of 3.0 m/s. a) How much of a resistive force up the ramp must there be? b) What is the kinetic coefficient of friction between the block and the ramp
Relevant Equations
Fy = 50cos30 = 43.3N
Fgx= 50 sin 30 = 25 N

a=Fnet/m
Fy = 50cos30 = 43.3N
Fgx= 50 sin 30 = 25 N

a=Fnet/m
3.0 = Fnet/5
Fnet=15

Ff = m Fn
15=m (43.3)
m=0.35
 
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  • #2
What is your question?
 
  • #3
Orodruin said:
What is your question?
Is part b right? .35 for kinetic coefficient of friction.
Or should it be
Ff = m Fn
(25-15) = m (43.3)
m=0.23
 
  • #4
galibe said:
Is part b right? .35 for kinetic coefficient of friction.
Or should it be
Ff = m Fn
(25-15) = m (43.3)
m=0.23
Which do you think, and why?
Think about the relationship between ##F_{net}, F_f ## and ##F_{gx}##.

Please don't use the same symbol (m) to mean different things. Write mu, or, better, use LaTeX.
 
  • #5
haruspex said:
Which do you think, and why?
Think about the relationship between ##F_{net}, F_f ## and ##F_{gx}##.

Please don't use the same symbol (m) to mean different things. Write mu, or, better, use LaTeX.
I am trying to help my daughter. I think it is:
Ff = mu *Fn
15 = mu * 43.3
mu=0.35

She believes it is the other answer.
Ff = mu*Fn
(25-15) = mu* 43.3
mu=0.23
 
  • #6
That doesn’t really answer the question of why you think your answer is correct.
 
  • #7
galibe said:
I am trying to help my daughter. I think it is:
Ff = mu *Fn
15 = mu * 43.3
mu=0.35

She believes it is the other answer.
Ff = mu*Fn
(25-15) = mu* 43.3
mu=0.23
Your daughter should be helping you!
 
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  • #8
Orodruin said:
That doesn’t really answer the question of why you think your answer is correct.
I don’t know, I haven’t done physics in many years.
 
  • #9
PeroK said:
Your daughter should be helping you!
So she is right? I have not done this in a long time and trying to relearn with her notes.
 
  • #10
galibe said:
I don’t know, I haven’t done physics in many years.
If in need of guidance, your daughter is very welcome here. It is better to have a direct connection to helpers than having a go-between that may misinterpret or misrepresent.
 
  • #11
galibe said:
So she is right? I have not done this in a long time and trying to relearn with her notes.
Yes, she is right. Fnet is the net force, not the friction force. The relation is Fnet = 25 N - Ff. (Always use units in equations!!)
 
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  • #12
Orodruin said:
If in need of guidance, your daughter is very welcome here. It is better to have a direct connection to helpers than having a go-between that may misinterpret or misrepresent.
I won’t misrepresent. By reading on the internet and her notes I am not sure if it should be 25-15 or 15. I would generally have her brother help but he is away at college.
 
  • #13
galibe said:
So she is right? I have not done this in a long time and trying to relearn with her notes.
Yes, she's right.
 
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  • #14
Orodruin said:
Yes, she is right. Fnet is the net force, not the friction force. The relation is Fnet = 25 N - Ff. (Always use units in equations!!)
Thank you, that was the piece I was unsure of. Fnet = 25N- Ff
I copied and pasted from the picture and some units didn’t come over.
 
  • #15
galibe said:
I won’t misrepresent.
Not consciously obviously. However, if the underlying understanding of the concepts is lacking, there is always a risk of misrepresentation. For example, you misrepresented the material from the book by assuming Ff = 15 N. I am sure that was not your intention, but it occurred.

We are here to help, but it is always easier to help the student directly.
 
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  • #16
I don't see any drawing of a free-body diagram (FBD). You may copy and paste your own sketch on pen and paper. Better still, there are excellent drawing softwares.
I think an FBD is essential for problems of this sort, specially if you're a beginner.
 
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  • #17
1706252556506.png
There you go @galibe, I draw for you the diagram of the problem using MSWord. You need to find the resistive force ##\color{red}{f_R}##. It took a little more than ten minutes.

Please note that I haven't shown all the forces. Can you complete the diagram by showing them?
 
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  • #18
galibe said:
I am trying to help my daughter. I think it is:
Ff = mu *Fn
15 = mu * 43.3
mu=0.35

She believes it is the other answer.
Ff = mu*Fn
(25-15) = mu* 43.3
mu=0.23
If you have some reasoning behind your answer, please state it. If you don't, why are you hampering your daughter?
 

FAQ: Solved: Help with a block sliding down an Inclined platform

How do I calculate the acceleration of a block sliding down an inclined plane?

To calculate the acceleration of a block sliding down an inclined plane, you can use the formula: \( a = g \sin(\theta) \), where \( g \) is the acceleration due to gravity (9.8 m/s²) and \( \theta \) is the angle of the incline. If there is friction, the formula becomes \( a = g (\sin(\theta) - \mu \cos(\theta)) \), where \( \mu \) is the coefficient of friction.

What forces are acting on a block sliding down an inclined plane?

The forces acting on a block sliding down an inclined plane include the gravitational force (weight) acting vertically downward, the normal force perpendicular to the surface of the incline, and the frictional force opposing the motion (if friction is present). The gravitational force can be split into two components: one parallel to the incline (causing the block to slide) and one perpendicular to the incline (contributing to the normal force).

How do I determine the coefficient of friction on an inclined plane?

The coefficient of friction (\( \mu \)) can be determined experimentally by measuring the angle (\( \theta \)) at which the block just starts to slide. At this angle, the force due to gravity parallel to the incline equals the maximum static friction force. The coefficient of friction is given by \( \mu = \tan(\theta) \).

How do I calculate the normal force on an inclined plane?

The normal force (\( N \)) on an inclined plane can be calculated using the formula: \( N = mg \cos(\theta) \), where \( m \) is the mass of the block, \( g \) is the acceleration due to gravity, and \( \theta \) is the angle of the incline. This force acts perpendicular to the surface of the incline.

How do I find the velocity of a block after sliding a certain distance down an inclined plane?

To find the velocity of a block after sliding a certain distance down an inclined plane, you can use the kinematic equation: \( v^2 = u^2 + 2ad \), where \( v \) is the final velocity, \( u \) is the initial velocity (which is often 0 if starting from rest), \( a \) is the acceleration, and \( d \) is the distance traveled. Solve for \( v \) to find the velocity.

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