Answer:
229.11 N
Explanation:
Mass of the shot (m) = 7.0 kg
Distance moved by the shot (S) = 2.2 m
Initial velocity of the shot (u) = 0 m/s (assume)
Final velocity of the shot (v) = 12 m/s
First we will determine the acceleration of the shot using the following equation of motion.
[tex]v^2=u^2+2aS\\\\(12)^2=0+2a(2.2)\\\\144=4.4a\\\\a=\frac{144}{4.4}=32.73\ m/s^2[/tex]
So, the acceleration of the shot is 32.73 m/s².
Now, using Newton's second law, the force acting on the shot is equal to the product of mass and acceleration. So,
Force = Mass × Acceleration
[tex]F=ma\\\\F=7\ kg \times 32.73\ m/s^2\\\\F=229.11\ N[/tex]
Therefore, the average force exerted by the shot-putter is 229.11 N.
