What are the values of the forces in question

The Gobbleblaster 3000 is the latest turkey cannon from Flying Fowl Industries. Measuring 3 meters in length it can launch a turkey from rest to a velocity of 100 m/s. Assuming the Gobblemaster 3000 is mounted on the floor, determine the following for a 25kg turkey.
a. The force the Gobbleblaster exerts on the turkey while it is in the cannon.
b. The normal force of the floor on the turkey.
c. Assuming a resistive frictional force (directed opposite the motion) of 200 N, how far away from the cannon does the turkey go?
d. What frictional force is required for the turkey to travel 1,250 meters from the cannon?
e. Draw a free body of the turkey from part d.

Hello William,

I am assuming that the turkey only slides across the floor, not fly through the air because of the resistive friction force.

a) 25 kg down

B) 25 N up

c) Sum of forces = ma, and the only force in the horizontal direction after the launch is Ff (friction)

Ff = m * a

a = Ff/m

and the equation of motion without time is

Vf^2 = Vi*2 + 2aX where x is the distance traveled.

Vf = zero when the turkey stops

Vi^2 = - 2aX

Solve for X

X = - Vi^2/2a

Substitute Ff/m for a

X = - Vi^2 * m / 2Ff

Plug in the numbers, Ff = - 200 N, Vi = 100 m/s, m = 25 kilos (holy moly, a 55 lb turkey??)

X = - (100 m/s)^2 * 25 kg / (2 * - 200 N)

X = 625 m

b) You can rearrange the equation and solve for Ff

Ff = Vi^2 * m / 2X

but we can see that they are inversely proportional, so for the distance to double from 625 m to 1250 m, then the Friction force must be halved to 100 N.

Good Luck