# How does $KE_f=KE_i+W$

I'm working on a problem of a ball with mass 76 sliding on a rough surface with the initial velocity of 4m/s. Friction does -100J of work and trying to find how much impulse did it impact? I worked with my professor on this problem and he used $$KE_f=KE_i+W$$ and then solved for $$v_f$$ since impulse $$J=m(v_f-v_i)$$. What I don't get is where he got the first equation from?

• Hello! It is preferable to use MathJax (LaTeX) to display formulas. You can find a tutorial at MathJax basic tutorial and quick reference. Please edit your question accordingly. Thanks! Dec 7, 2021 at 23:17

What I don't get is where he got the first equation from?

He got it from the work energy theorem which states that the net work done on an object equals its change in kinetic energy. The equation

$$KE_f=KE_i+W$$

Can be re-written as

$$KE_{f}-KE_{i}=W_{net}$$

and for translational kinetic energy

$$\frac{1}{2}mv_{f}^{2}-\frac{1}{2}mv_{i}^{2}=W_{net}$$

Where subscripts $$i$$ and $$f$$ indicate initial and final velocities.

I'm assuming you want to know how we get this equation.

We can derive the equation by combining the laws of motion, Newton's second law, and the equation for work as follows:

From kinematics (equations of motion) we get, for constant acceleration

$$v_{f}^{2}=v_{i}^{2}+2ad$$

$$v_{f}^{2}-v_{i}^{2}=2ad$$

where $$a$$ is a acceleration of the object due to a net force and $$d$$ is the displacement of the object.

By multiplying both sides of the equation by $$m/2$$ and expressing acceleration and displacement as vector quantities:

$$\frac{1}{2}mv^{2}_{f}-\frac{1}{2}mv^{2}_{i}=m\vec a.\vec d$$

From Newton’s second law

$$\vec F_{net}=m\vec a$$

substituting into the previous equation gives

$$\frac{1}{2}mv^{2}_{f}-\frac{1}{2}mv^{2}_{i}=\vec F_{net}.\vec d$$

And finally, since $$\vec F_{net}.\vec d$$ equals the net work done, $$W_{net}$$, we get the equation for the work energy theorem

$$\frac{1}{2}mv^{2}_{f}-\frac{1}{2}mv^{2}_{i}=W_{net}$$

Hope this helps.

The work - energy theorem defines kinetic energy to be so, I will highly recommend to go through that discussion either in your book or over the Internet.