Newton's Second Law problem

[rasen58]

A 2.0 kg otter starts from rest at the top of an incline 85 cm long and slides down to the bottom in 0.50 seconds. What net external force acts on the otter along the incline?

 

So I was planning on using F=ma, and since we have the mass, we need the acceleration to find F.

 

a = change in velocity/time.

velocity = change in displacement/time = 85/.5 = 170 cm/s =1.7 m/s

a = 1.7/.5 = 3.4 m/s²

 

Then F=(2)(3.4) = 6.8 N. But that is wrong because it should be 14 N instead.

 

Apparently if you use delta x = v₀t + .5at², you can solve for a that way and you get

a = 2(delta x)/(t²) = (2)(.85)/(.5²) which gives you 6.8 m/s²

And then if you use F=ma, you get (2)(6.8)=13.6=14 N.

 

Why did my first way not work?

[imatfaal]

A 2.0 kg otter starts from rest at the top of an incline 85 cm long and slides down to the bottom in 0.50 seconds. What net external force acts on the otter along the incline?

 

So I was planning on using F=ma, and since we have the mass, we need the acceleration to find F.

 

a = change in velocity/time.

velocity = change in displacement/time = 85/.5 = 170 cm/s =1.7 m/s

a = 1.7/.5 = 3.4 m/s²

 

Then F=(2)(3.4) = 6.8 N. But that is wrong because it should be 14 N instead.

 

Apparently if you use delta x = v₀t + .5at², you can solve for a that way and you get

a = 2(delta x)/(t²) = (2)(.85)/(.5²) which gives you 6.8 m/s²

And then if you use F=ma, you get (2)(6.8)=13.6=14 N.

 

Why did my first way not work?

 

 

You are getting confused between average velocity and final velocity. In the simple formulation (constant a) acceleration is the change in velocity from start to finish divided by the time - you have used the difference between the average velocity and starting velocity divided by the time.

 

I would draw a velocity time diagram - you will know that the displacement is the area under the line and acceleration is the slope. Diagrams are a very useful way of finding a second approach when your first is stuck - they are the same physics but rely on different bits of knowledge and you are unlikely to get tripped up in two different ways

[studiot]

 

Apparently if you use delta x = v₀t + .5at², you can solve for a that way and you get

a = 2(delta x)/(t²) = (2)(.85)/(.5²) which gives you 6.8 m/s²

And then if you use F=ma, you get (2)(6.8)=13.6=14 N.

 

A simple distinction.

 

The study of how fast, how far and for how long bodies travel, without worrying about the forces involved, is often called kinematics.

 

The study of motion including forces, energies etc is called dynamics.

 

Your problem above links the two together.

 

Kinematics includes equations such as distance = speed x time and the more complicated one above that you mention.

Have you studied kinematics yet?

[swansont]

 

Kinematics includes equations such as distance = speed x time and the more complicated one above that you mention.

Have you studied kinematics yet?

 

And if you haven't, can you derive x = v₀t + .5at² from the definitions of acceleration and velocity?

[rasen58]

Oh wow, that makes sense, thank you.

 

Swansont, yes I have derived them multiple times before when I used to forget it.