How Does Work Affect Acceleration on an Air Track?

[msimard8]

Here is the queston.

A student in a physics lab pushed a 0.100 kg cart on an air track over a distance of 10.0 cm, doing 0.230 J of work. Calculate the acceleration of the cart.

The textbook doesn't show how to approach this problem. It is suppose to be bonus or somthing.

These are my thoughts

I am assuming there is no friction since the cart was on a air track

I converted everything to their proper units.

mass=0.100kg
w=0.0230 J
d =10.0 cm = 0.1m

I know accleration is change of velocity/time

I need a hint at least to figure out how to solve this.

Thanks

 

[FredGarvin]

First, think of acceleration in terms of Newton's second law...

You have been given the amount of work done on the cart and the distance that work was performed. What is the definition of work? What is left to calculate from that definition, i.e. you have been given two things and there are three things in the basic definition for work.

Also, you have one too many leading zeros in the work number when you showed converting to proper units.

 

[quicknote]

for your question. The first step in solving this problem is to use the work-energy theorem, which states that the work done on an object is equal to the change in kinetic energy of the object. In this case, the work done on the cart is equal to the change in kinetic energy of the cart. We can express this mathematically as:

W = ΔKE = 1/2mv^2 - 1/2mv0^2

Where W is the work done, m is the mass of the cart, v is the final velocity of the cart, and v0 is the initial velocity of the cart (which we can assume is 0 since the cart starts from rest).

Next, we can substitute the given values into this equation:

0.0230 J = 1/2(0.100 kg)(v^2) - 1/2(0.100 kg)(0 m/s)^2

Simplifying, we get:

0.0230 J = 1/2(0.100 kg)(v^2)

Now, we can solve for v by rearranging the equation:

v = √(0.046 J/kg)

Finally, we can use the definition of acceleration (a = Δv/Δt) to find the acceleration of the cart. Since the cart traveled a distance of 10 cm (0.1 m) in a time of t, we can write:

a = v/t = √(0.046 J/kg) / t

We can't determine the exact time t from the given information, but we can use the distance and the fact that the cart is initially at rest to find an approximate value for t. We can use the equation d = 1/2at^2, where d is the distance, a is the acceleration, and t is the time. Rearranging this equation, we get:

t = √(2d/a)

Substituting in the given values, we get:

t = √(2 * 0.1 m / √(0.046 J/kg))

Simplifying, we get:

t = √(4.35 s^2)

Therefore, the approximate value of t is 2.08 seconds.

Plugging this value into our equation for acceleration, we get:

a = √(0.046 J/kg) / 2.08 s

Simplifying,