Calculating Net Force on 2kg Mass Down 8m Incline

In summary, The problem involves a 2kg mass sliding down an inclined plane with a length of 8.0x10m in 0.50s. To find the net force acting on the mass, we need to use a motion formula to find the acceleration. It is suggested to draw a free body diagram and sum the forces on the body. The solution involves using the kinematic equation d=vi*t+0.5*a*t^2, with the initial velocity being zero. The distance traveled is 8.0x10m.
  • #1
NickS606
9
0

Homework Statement



A 2kg mass starts from rest and slides down an inclined plane 8.0X10m long in 0.50s. What net force is acting on the mass along the incline?

Homework Equations





The Attempt at a Solution



I don't even know where to begin.
 
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  • #2
This is an accelerated motion problem. Use a motion formula to find the acceleration.
 
  • #3
Try drawing a free body diagram and then sum the forces on the body.
 
  • #4
I'm sorry I am still new, can I get a better explanation please?
 
  • #5
NickS606 said:
I'm sorry I am still new, can I get a better explanation please?
Can you work out the acceleration of the object?
 
  • #6
How can I do that?
 
  • #7
NickS606 said:
How can I do that?
Do you know any Kinematic (SUVAT) equations?
 
  • #8
I'm just going to go ahead and say I have never even heard of those equations.
 
  • #9
NickS606 said:
I'm just going to go ahead and say I have never even heard of those equations.
Really? Then I'm going to go ahead and say take a look in your textbook or class notes and find them.

Alternatively, you can find them https://www.physicsforums.com/showpost.php?p=905663&postcount=2". However, I must say that I will be extremely surprised if your teacher hasn't mention them, but set you a homework that requires them.
 
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  • #10
Ya I don't really recall those. Anyways, would I do [tex]a_{ave} = \Delta v / \Delta t[/tex]

I did 12N/3.5s and my answer is 3.4, is this correct?
 
  • #11
NickS606 said:
Ya I don't really recall those. Anyways, would I do [tex]a_{ave} = \Delta v / \Delta t[/tex]
Perhaps, but we'll have to do some work first.
NickS606 said:
I did 12N/3.5s and my answer is 3.4, is this correct?
No. Did you just pluck those numbers out of thin air?

Returning to the question, what do the dimensions 8x10m refer to? Height and width or height and hypotenuse or width and hypotenuse? Do you have a diagram?
 
  • #12
I got the numbers out of the question, and there is no diagram, it is height and width for sure though.
 
  • #13
NickS606 said:
I got the numbers out of the question, and there is no diagram, it is height and width for sure though.
Okay. Have you looked for the kinematic equations in your notes or text?
 
  • #14
Well there aren't many notes and all I could find was this equation [tex]v_{ave} = \Delta x / \Delta t[/tex]

The teacher isn't really that great and won't help much, that's why I am here.
 
  • #15
NickS606 said:
Well there aren't many notes and all I could find was this equation [tex]v_{ave} = \Delta x / \Delta t[/tex]
Okay, well we cannot use that equation. Try reading http://www.glenbrook.k12.il.us/GBSSCI/PHYS/Class/1DKin/U1L6a.html" and see if it jogs your memory.
 
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  • #16
I'm just not getting it, I assume the equation to be used is

d=vi*t+1.5*a*t^2
 
  • #17
NickS606 said:
I'm just not getting it, I assume the equation to be used is

d=vi*t+1.5*a*t^2
The highlighted coefficient should be 0.5 rather than 1.5, but yes, this is the equation to use. Can you start filling in the known variables?
 
  • #18
Yes, that formula will do it if the initial velocity is zero.
 
  • #19
Is it d=0*.50s+.5*a*.50s^2

I put a for acceleration because I don't know what acceleration is :/
 
  • #20
Delphi51 said:
Yes, that formula will do it if the initial velocity is zero.
One may use the above formula even if the initial velocity is non-zero.
NickS606 said:
Is it d=0*.50s+.5*a*.50s^2

I put a for acceleration because I don't know what acceleration is :/
Acceleration is what you want to find out! You need to solve for the acceleration.

Now, what about the distance traveled?
 

What is the formula for calculating net force on a 2kg mass down an 8m incline?

The formula for calculating net force on a 2kg mass down an 8m incline is: Fnet = mgsin(theta), where Fnet is the net force, m is the mass, g is the acceleration due to gravity (9.8 m/s^2), and theta is the angle of the incline.

What is the unit for net force?

The unit for net force is Newtons (N).

How do you calculate the angle of the incline?

The angle of the incline can be calculated using the formula: tan(theta) = opposite/adjacent, where theta is the angle, opposite is the height of the incline, and adjacent is the length of the incline.

What is the significance of calculating net force on a mass down an incline?

Calculating net force on a mass down an incline helps determine the acceleration of the mass. This can be used to understand the motion of objects on an incline and to make predictions about their behavior.

Can the net force on a mass down an incline ever be zero?

Yes, the net force can be zero if the mass is at rest or if the forces acting on the mass are balanced. This typically occurs when the incline is at a certain angle, known as the equilibrium angle, where the forces pulling the mass down the incline and the forces pushing the mass up the incline are equal.

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