DYNAMICS:
A 54kg fireman slides down a pole of a distance of 3m. if the man hits the groung at 4.4m/s than what will the force of friction acting on the man due to the pole?
To determine the force of friction acting on the fireman due to the pole, we can use the principles of dynamics.
1. First, we need to calculate the initial potential energy of the fireman at the top of the pole. Since potential energy is given by the formula PE = mgh, where m is the mass (54kg), g is the acceleration due to gravity (9.8m/s^2), and h is the height (3m), we have PE = 54kg * 9.8m/s^2 * 3m = 1587.6 Joules.
2. Next, we need to calculate the final kinetic energy of the fireman when he hits the ground. Kinetic energy is given by the formula KE = 1/2mv^2, where v is the final velocity (4.4m/s). Therefore, KE = 1/2 * 54kg * (4.4m/s)^2 = 529.52 Joules.
3. The difference between the initial potential energy and the final kinetic energy represents the energy lost due to friction. So, the work done by friction is given by W = PE - KE = 1587.6J - 529.52J = 1058.08 Joules.
4. Finally, the force of friction can be obtained by dividing the work done by friction by the distance the fireman slides down the pole. Therefore, the force of friction is F = W / d = 1058.08J / 3m = 352.7 Newtons.
So, the force of friction acting on the fireman due to the pole is approximately 352.7 Newtons.