Question

How much work does friction do if an 85-kilogram base runner travels at an initial velocity of 8 meters per second and slides into 2 nd base and stops? m=85kg v_(1)=8(m)/(s) v_(f)=0(m)/(s) W_(friction )

Answer 1

The work done by friction as an 85-kilogram base runner slides to a stop from an initial velocity of 8 meters per second is -2720 Joules.

Step-by-step explanation:

To determine how much work friction does as an 85-kilogram base runner, who travels at an initial velocity of 8 meters per second and slides into second base and stops, we can use the work-energy principle. The work done by friction is equal to the change in the kinetic energy of the base runner.

The initial kinetic energy (KEi) of the runner is given by the formula KEi = 1/2 mvi2, where m is the mass and vi is the initial velocity. Using the given values, KEi=1/2 x 85 kg x (8 m/s)2 = 2720 Joules. Since the final velocity (vf) is 0 meters per second, the final kinetic energy (KEf) is 0 Joules.

According to the work-energy principle, the work done by friction will be equal to the change in kinetic energy, which is KEf - KEi. Hence, the work done by friction Wfriction is -2720 Joules (negative because friction does work in the direction opposite to motion).