Calculating Net Force on an Otter Using Newton's Second Law

AI Thread Summary
To calculate the net force acting on a 2.0 kg otter sliding down an 85 cm incline in 5.0 seconds, one must first determine the acceleration using the kinematic equation. The otter starts from rest, meaning initial velocity is 0 m/s, and the incline's slope should be considered as the x-axis for calculations. The confusion arises from misunderstanding the forces at play; unlike projectile motion, the incline exerts a force on the otter, resulting in acceleration down the slope. Ultimately, the net force can be calculated as 14.0 N, confirming that the otter does experience acceleration along the incline. Understanding the distinction between projectile motion and motion on an incline is crucial for solving such problems.
apples
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1. A 2.0 kg otter starts from rest at the top of a muddy incline 85 cm long and slides down to the bottom in 5.0 s. What net force acts on the otter along the incline.



2. Homework Equations - F=ma, maybe some more too.



3. The Attempt at a Solution
I attempted this question for one and a half hour, but I couldn't get it. This is coming in my exam TOMORROW, but I just don't get it.
The answer is 14.0 N, but I just don't know how to get it. And it's too late to ask help from a school teacher. I've done this question before, but someone took my physics copy to copy all the assignments and never gave it back...
Now I don't know what to do.
 
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you can calculate the acceleration along the incline using

x-x_i = v_i t + \frac{at^2}{2}

x_i is the initial position along the incline : 0 m
v_i is the initial velocity along the incline : 0 m/s

marlon
 
The initial x position isn't 0 m, it's an incline :S
 
apples said:
The initial x position isn't 0 m, it's an incline :S
You are looking for the acceleration and the net force in the direction of the incline; therefore, you can choose the slope of the incline to be the x axis, and set x_initial = 0, as previously noted. Also, recheck the value of the correct answer.
 
apples said:
The initial x position isn't 0 m, it's an incline :S

Yes it is. Check how i defined the reference frame (Like PhantomJay explains).

Again just take the origin at the starting point. The x-axis is along the incline. You can chose the X and Y axis anyway you want.

marlon
 
PhanthomJay said:
Also, recheck the value of the correct answer.
Indeed

marlon
 
I still don't get it
Suppose i take the incline as the x axis...
xf-xi=.85m
but vi=0 m/s
and the acceleration in the x direction is 0 m/s... or not??
This is the part I'm getting confused in
There's no accelration in the x direction.
Suppose i take the condition as it is.
According to my frame of reference the height of the incline is on the y axis.
the incline is a horizontal line with a negative slope...
then the acceleration in the x direction is 0 so is vi...
I don't understand. how can 0=0.85...

can someone please explain!
 
apples said:
I still don't get it
Suppose i take the incline as the x axis...
xf-xi=.85m
but vi=0 m/s
and the acceleration in the x direction is 0 m/s... or not??
This is the part I'm getting confused in
There's no accelration in the x direction.

What makes you think there is no acceleration in the x direction? You are looking for the net force acting on the otter.. this force will provide the acceleration. If you consider, like marlon says, taking x_i = 0 and v_i = 0, then there must be an acceleration in order for the otter to move!

Using Marlon's equation

marlon said:
you can calculate the acceleration along the incline using

x-x_i = v_i t + \frac{at^2}{2}

x_i is the initial position along the incline : 0 m
v_i is the initial velocity along the incline : 0 m/s

marlon

work out the acceleration. Then you should be able to find the net force
 
apples said:
I still don't get it
Suppose i take the incline as the x axis...
xf-xi=.85m
but vi=0 m/s
and the acceleration in the x direction is 0 m/s... or not??
Why in the world would you think there's no acceleration down the incline? If there were no acceleration, the otter would just sit there. But he slides down!

Taking x to be along the incline, use the given kinematics equation to solve for the acceleration.
 
  • #10
I think I'm confusing it a bit with projectile motions...
i was thinking that that the otter slid down the incline because of gravity. and since there is an incline, he slides down it, but... errr.. I don't know, I think I"m confusing myself. Let me see if I can solve this now.
EDIT: can someone explain why the acceleration in a projectile motion without air resistance is zero, and why the acceleration here isn't zero?
 
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  • #11
apples said:
can someone explain why the acceleration in a projectile motion without air resistance is zero, and why the acceleration here isn't zero?
The acceleration of a projectile without air resistance is 9.8 m/s^2 down. Not zero!

The only force on a free projectile is gravity, and that acts in such a way that all objects experience the same acceleration--called the acceleration due to gravity. :wink:
 
  • #12
Yeah so the accelration along the y-axis is 9.81 m/s^2, and the acceleration along the x-axis is 0!...
So why wasn't the only force acting on the otter here gravity? Please explain, I hope this answer would remove all my confusions.
Thanks a lot
 
  • #13
apples said:
Yeah so the accelration along the y-axis is 9.81 m/s^2, and the acceleration along the x-axis is 0!...
For a free projectile, where the y-axis is vertical and the x-axis is horizontal, this is correct.

But in your problem:
(1) Gravity is not the only force acting on the otter. The incline exerts a force on the otter! That makes a big difference: something sliding down a ramp is not a projectile.
(2) The x-axis is not horizontal, it is parallel to the incline. This is done for convenience, since we know that the only motion will be parallel to the incline.
 
  • #14
Thanks a lot! I think I'm clear about all this now. :)
 
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