HELP, Physics homework!

Question

Ugh, I'm having hard time dealing with numbers. I know these questions were very simple, but (desperate mode), I'm in need of your help guys.

So, here's the problem:

When a 58-g tennis ball is served, it accelerates from rest to a speed of 45m/s. The impact with the racket gives the ball a constant acceleration over a distance of 44cm. What is the magnitude of the net force acting on the ball?

First one to answer will get 10 points. :)

Thanks for reading this.

Answer 1

Well,

F = m*a   where m is the mass, and a is the acceleration.

We have the mass which is 58g or .058 kg

The acceleration can be solved using the following equation...

xf = xo + vt + .5at^2

where

xf = .44cm

vo = 0

vf = 45 m/s

a = (vf - v0) / t

.44 = 0 + .5(45)*t

t = 0.01956

So the acceleration is a = 45m/s / 0.01956 s

a = 2301 m/s^2

The force acting on the ball due to the racket is ...

F = .058kg * 2301 m/s^2

F = 133.47 N

2)

Since no energy is lost to friction, you can use the following equation...

KE1 + PE1 = KE2

.5*m*v_1^2 + mgh = .5*m*v_2^2

where v_1 is the initial velocity, and v_2 is the velocity at the bottom of the ramp.

The height h can be found using the angle and length of the ramp.

h = 6*sin(18) = 1.854 m

so...

.5*2.6^2 + 9.81*1.854 = .5*v_2^2

3.38 J + 18.187 J = .5*v_2^2

v_2 = 6.567 m/s

Answer 2

v^2=u^2 + 2 fs

or final velocity squared = initial velocity squared + twice the acceleration times the distance through which it acts.  Substitute and you will get the answer but don't ask me what the answer is in as I learned in MKS units and don't even think to ask what MKS units are!