A high-speed photograph of a club hitting a golf ball is shown in the figure below. The club was in contact with a ball, initially at rest, for about 0.0013 s. If the ball has a mass of 55 g and leaves the head of the club with a speed of 1.2 102 ft/s, find the average force exerted on the ball by the club.
F=mv/t
To find the average force exerted on the ball by the club, we can use Newton's second law of motion, which states that force is equal to mass times acceleration. In this case, the acceleration can be calculated using the equation for average acceleration:
average acceleration = (final velocity - initial velocity) / time
First, we need to convert the velocity of the ball from feet per second to meters per second:
1 ft/s = 0.3048 m/s (conversion factor)
So, the final velocity of the ball is:
1.2 x 10^2 ft/s * 0.3048 m/s/ft = 36.576 m/s
The initial velocity of the ball is 0 m/s since it was initially at rest.
Now, we can calculate the average acceleration:
average acceleration = (36.576 m/s - 0 m/s) / 0.0013 s = 28135.38 m/s^2
Next, we convert the mass of the ball from grams to kilograms:
55 g = 0.055 kg (conversion factor)
Now we can calculate the average force exerted on the ball:
average force = mass * acceleration = 0.055 kg * 28135.38 m/s^2 = 1547.4459 N
Therefore, the average force exerted on the ball by the club is approximately 1547.45 Newtons.