Estimate the average force exerted by a shot-putter on a 6.5 kg shot if the shot is moved through a distance of 2.7 m and is released with a speed of 11 m/s.
Final kinetic energy = F*X
(1/2)MV^2 = 2.7 meters * F
Solve for F
To estimate the average force exerted by a shot-putter, we can use Newton's second law which states that the force exerted on an object is equal to the mass of the object multiplied by its acceleration. In this case, the acceleration is related to the change in velocity over time.
First, let's calculate the initial kinetic energy of the shot. The kinetic energy (KE) formula is given by KE = 1/2 * mass * velocity^2.
The mass of the shot is given as 6.5 kg, and the velocity is given as 11 m/s. Substituting these values into the formula:
KE = 1/2 * 6.5 kg * (11 m/s)^2
= 1/2 * 6.5 kg * 121 m^2/s^2
= 394.25 J
The initial kinetic energy of the shot is 394.25 Joules.
Now, we need to calculate the final kinetic energy of the shot after traveling a distance of 2.7 m. Since we know the initial and final velocities are the same (0 m/s because the shot is released), we can use the work-energy principle, which states that the work done on an object is equal to its change in kinetic energy.
The work done (W) on the shot is given by W = F * d, where F is the average force exerted on the shot and d is the distance traveled.
Using the formula for work, we have:
W = ΔKE
F * d = KE_final - KE_initial
Substituting the values we know:
F * 2.7 m = 0 J - 394.25 J
Simplifying:
F = -394.25 J / 2.7 m
≈ -146 J/m
The average force exerted by the shot-putter on the 6.5 kg shot is approximately -146 J/m. Note that the negative sign indicates that the direction of the force is opposite to the direction of the displacement.