How Much Force Does a Golf Club Apply to a Ball?

[PhysicFailure]

Homework Statement

When a 0.045 kg golf ball takes off after being hit, its speed is 42 m/s.
(a) How much work is done on the ball by the club?

---> 39.69 J
(b) Assume that the force of the golf club acts parallel to the motion of the ball and that the club is in contact with the ball for a distance of 0.030 m. Ignore the weight of the ball and determine the average force applied to the ball by the club.

Homework Equations

W_NCF= [tex]\DeltaME_TOTAL[/tex]

The Attempt at a Solution

for part a I did this and got the correct answer in Joules: 0.5(0.045kg)(42²

I am confused at what to do in the next part. I did the same attempt but confused when I am told to ignore the weight of the ball...

 

[PhysicFailure]

2. Homework Equations

W_ncf= $$\Delta$$ME_total

 

[HallsofIvy]

Homework Statement

When a 0.045 kg golf ball takes off after being hit, its speed is 42 m/s.
(a) How much work is done on the ball by the club?

---> 39.69 J
(b) Assume that the force of the golf club acts parallel to the motion of the ball and that the club is in contact with the ball for a distance of 0.030 m. Ignore the weight of the ball and determine the average force applied to the ball by the club.

Homework Equations

W_NCF= [tex]\DeltaME_TOTAL[/tex]

The Attempt at a Solution

for part a I did this and got the correct answer in Joules: 0.5(0.045kg)(42²

I am confused at what to do in the next part. I did the same attempt but confused when I am told to ignore the weight of the ball...

Okay: the kinetic energy of the ball changed from 0 to (1/2) mv². That energy had to come from somewhere- the work done on it.

Now do you know that when on object accelerates from 0 with constant (or average) acceleration a to t seconds it reachs speed a*t and goes a distance (1/2)at². You know that the ball accelerated to 42 m/s while going 0.30m. That gives two equations in a and t. Once know a, the average acceleration, F= ma gives the average force.