A tennis ball dropped vertically (Newton's Forces)

1. Nov 14, 2009

Chandasouk

1. The problem statement, all variables and given/known data

A tennis ball has a mass of 0.057 kg. After falling 18 m vertically from rest it has a speed of 12 m/s. What is the magnitude of the average resistive force acting on it as it falls?

My answer for this was .3306N

mass = 0.057kg

Vi = 0 m/s

Vf = 12m/s

Xi = 18m

Xf=0m

I started with w=mg = -0.05586N

So FN must be 0.05586N since it is dropped vertically. The free body diagram would be FN pointed up and weight pointing straight down.

I used the equation V^2 = Vi^2 +2a$$\Delta$$x

(12m/s)^2 = 2a(-18m)

a = -4m/s^2

F=ma

F=(.057kg)(-4m/s^2) = -0.228N

Fnet = Fn + Fresistance = .5586N + (-.228N) = .3306N

2. Nov 14, 2009

Delphi51

Looks good!

3. Nov 14, 2009

mg0stisha

What is the normal force due to in this problem? Wouldn't there only be the force of gravity (pointing down) and the resistive force (pointing up)?

4. Nov 14, 2009

Delphi51

There isn't a normal force. Yes, that solution is confusing. Better to write
mg - F = ma (downward is positive)
F = m(g-a) = .057(g - 4) = 0.331 N.

5. Nov 14, 2009

mg0stisha

Okay, thank you. I recognized that after i replied/worked the solution out myself.

6. Nov 15, 2009

Chandasouk

So you set the Fy which is mg and F equal to ma? What would the equation look like if you said downward was negative? I worked it out and got the acceleration as negative. Is it because you said downward was positive that -4m/s^2 was now just 4m/s^2?

Last edited: Nov 15, 2009