The alarm at a fire station rings and an 86-kg fireman, starting from rest, slides down a pole to the floor below (a distance of 4.0 m). Just before landing, his speed is 1.4 m/s. What is the magnitude of the kinetic frictional force exerted on the fireman as he slides down the pole?
we begin this question by calculating the acceleration off the fire man. We can do that by using. Torricelli's equation tells us that the final velocity squared is interested. Initial velocity squared plus two times his acceleration times. He's displacement. Also, let me choose that everything that is pointing upwards is positive. Then his final velocity is 1.4 meters per second, his initial velocity zero because he starts from rest, then we have plus two times the acceleration times they displacement on displacement Waas four meters but four meters downwards. Therefore, he started at some height and finishes at his water height, meaning that in my reference frame the displacement is negative and equals two minus four liters. Then we solve this equation for eight as follows. 1.4 squared is the goes to minus eight times a so that his acceleration is equals to 1.4 squared, divided by eight with a minus sign in front of it on. These is minus 0.245 meters per second squared. Now that you know his acceleration, we can use Newton's second law to calculate what is the frictional force. Newton's second law tells us that the net force is the cost of the mass times acceleration off the fireman. Then the net force is composed by true forces, the frictional force that points of ports. So it's positive minus the weight force that points downwards under four negative. And these Zico's to the Mass 86 kilograms times acceleration off minus zero point True for five meters per second squared. Then the frictional force is he goes to 86 times, minus 0.2 for five. Plus he's wait. Remember that the weight is given by the mass times acceleration off gravity, which is approximately 9.8 meters per second, squared near the surface of the earth. So plus 86 times 9.8 then the frictional force is approximately 820 new tons.