Question

#1 A 69.6 kg runner has a speed of 3.30 m/s at one instant during a long-distance event.

If the runner's kinetic energy at this instant is 378.98J, how much net work (in J) is required to double his speed?

2b. A 52.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 130 m/s from the top of a cliff 156 m above level ground, where the ground is taken to be y = 0. The initial total mechanical energy of the projectile = 518897.6; Suppose the projectile is traveling 92.1 m/s at its maximum height of y = 332 m. How much work has been done on the projectile by air friction?

2c. What is the speed of the projectile immediately before it hits the ground if air friction does one and a half times as much work on the projectile when it is going down as it did when it was going up?

Answer 1

1] Net KE required = 0.5mv^2 = 0.5*69.6*(2*3.3)^2 - 378.98 = 1136.9 J

2a] work done by all forces = change in KE

work done gravity + work done by air friction = 0.5mv^2 - 0.5mu^2

mg* delta h + work done by friction = 0.5mv^2 - 0.5mu^2

52*9.8*(156- 332) + work done by friction = 0.5*52*92.1^2- 0.5*52*130^2

work done by friction = 0.5*52*92.1^2- 0.5*52*130^2 - 52*9.8*(156- 332)

= -129167.7 J

2b) work done by all forces = change in KE

   work done gravity + work done by air friction = 0.5mv^2 - 0.5mu^2

   52*9.8*(156- 0) -129167.7*(1+1.5) = 0.5*52*v^2- 0.5*52*130^2

v = 86.8 m/s answer