What net force is acting on the object along the incline?

AI Thread Summary
An object of mass 2.0 kg slides down a 69 cm incline in 0.90 seconds, prompting a calculation of the net force acting on it. The initial calculation of acceleration was incorrect due to a misapplication of the formula, specifically in the algebraic manipulation. The correct approach involves using the equation deltaX = 1/2at^2, leading to the acceleration being calculated as a = 2 * deltaX / t^2. This adjustment reveals that the net force can be accurately determined using Fnet = ma. The discussion highlights the importance of careful algebra in physics problem-solving.
rculley1970
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Problem:

An object of mass 2.0kg starts from rest and slides down an inclined plane 69cm long in 0.90s. What net force is acting on the object along the incline?


Now, I used what I was given and plugged it in the equation:

deltaX=1/2at^2 + V0t

I come up with acceleration = .0552 m/s^2

I plug this into equation:

Fnet = ma for a net force of .11N

This is wrong. Am I missing something? I don't have angle theta to use the equation: mgsin(theta).
 
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rculley1970 said:
Problem:

An object of mass 2.0kg starts from rest and slides down an inclined plane 69cm long in 0.90s. What net force is acting on the object along the incline?


Now, I used what I was given and plugged it in the equation:

deltaX=1/2at^2 + V0t

I come up with acceleration = .0552 m/s^2

I plug this into equation:

Fnet = ma for a net force of .11N

This is wrong. Am I missing something? I don't have angle theta to use the equation: mgsin(theta).


What did you do to get that for an acceleration, that's not what I get when I use your equation.
 
rculley1970 said:

Now, I used what I was given and plugged it in the equation:

deltaX=1/2at^2 + V0t

I come up with acceleration = .0552 m/s^2

Your thinking is correct, the only thing is that the acceleration you've calculated isn't the correct value.

The problem indicates that:

x= 69 cm = 69 * 10-2 m
v0= 0 m/s
t= 0.90 s

Because v0= 0 m/s the equation deltaX=1/2at^2 + V0t becomes:

deltaX=1/2at^2

therefore

2 delta X= at2 so
a= 2 delta X/ t2

Can you get to the right answer now?
 
oops, i messed up the easy algebra and put .5t^2 on the denominator instead of multiplying deltaX by two. Such an easy mistake can kill a problem. Thanks for everyones help.
 
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