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Second Law of Motion

So what happens if the forces on an object are not balanced? If the net force on an object is nonzero, then the velocity vector changes. There must be acceleration. In fact the larger the force, the larger the acceleration. (See Figure 3-5.) On the other hand, if we apply the same push to both a small car and a large car (Figure 3-6), the small car will have the larger acceleration.
 

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Figure 3-5

 

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Figure 3-6

 

We can write (in one dimension)
 

Image (1)
 

We have not proven this equation, but the discussion in the previous paragraph should make it seem reasonable to you.

In three dimensions, we write what is often called Newton’s second law:
 
Second Law of Motion
If an object has forces  …, acting on it, then the total force on the object is the vector sum
 
 
And the acceleration  of the object is given by
 
  (2)

 

Most often, however, we will break up equation (2) into components:
 

(3a)

 

Image (3b)
 

Equation (3a), for example, states that the sum of all the horizontal forces is mass times the horizontal acceleration. We will discuss breaking vectors into vertical and horizontal components in Section 4.D.
 
Finally, we are able to make the connection between the units for force [N] and for mass [kg], introduced in Chapter 1. The Newton is defined by
 

Image
 

Example

Bruce pushes a car (500 kg) on level ground starting from rest with a force 100 N. How long does it take to get the car rolling 1 m/s? (Assume no friction.)

Solution

We have the information m = 500 kg, v1 = 0 m/s, F = 100 N, and v2 = 1 m/s, and we want Δt. We can find acceleration from equation (1), so we obtain

 

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Then we can find Δt from

 

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Example

A rocket provides a constant force to wagon A that rolls without friction. It starts from rest and after time t attains velocity v. A similar rocket providing the same force is attached to wagon B, which has five times the mass of A (Figure 3-7). How much time does it take wagon B to go from rest to velocity v? (Assume no friction.)

 

Image

 
Solution

This one looks difficult, but if we write down the relevant equations, it is not so hard. We need to connect force and velocity, so we write,

 

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We set v1 to zero because the wagons start from rest. Substitution gives

 

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Since the problem asks about the change in time, we can solve for Δt to obtain

 

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Now, F and v2 stay the same, but m is five times larger for wagon B, so Δt is five times larger.
The answer is 5t.
 

 





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