A golf ball of mass 50g is accelerated from rest to 40m/s in 2ms - what is the average force exerted by the club on the ball?

a = 40 m/s / 2 ms = 20000 m/s^2

f = m a = .05 kg * 20000 m/s^2

answer is in Newtons

To calculate the average force exerted by the club on the golf ball, we can use Newton's second law of motion, which states that force (F) is equal to the mass (m) of an object multiplied by its acceleration (a): F = m * a.

First, we need to convert the mass of the golf ball from grams to kilograms since the unit for force is Newtons (N). We can do this by dividing the mass (50g) by 1000: 50g / 1000 = 0.05kg.

Next, we need to calculate the acceleration of the golf ball. Given that the ball is accelerated from rest (initial velocity, vi = 0m/s) to a final velocity, vf = 40m/s, in a time interval, t = 2ms = 0.002s, we can use the formula: vf = vi + a * t. Since the initial velocity is zero, the formula simplifies to: vf = a * t.

Rearranging the formula, we can solve for acceleration (a): a = vf / t.

Substituting the given values: a = 40m/s / 0.002s = 20,000 m/s^2.

Now that we have the mass (0.05kg) and the acceleration (20,000 m/s^2), we can calculate the average force exerted by the club on the golf ball using Newton's second law: F = m * a.

Substituting the values: F = 0.05kg * 20,000 m/s^2 = 1000 N.

Therefore, the average force exerted by the club on the golf ball is 1000 Newtons.