A tennis ball dropped vertically (Newton's Forces)

[Chandasouk]

Homework Statement

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[tex]\Delta[/tex]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

 

[Delphi51]

Looks good!

 

[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)?

 

[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.

 

[mg0stisha]

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

 

[Chandasouk]

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.

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?