A 5kg ball is thrown upward with an initial speed of 30m/s. Suppose there is a constant air resistance of 25N opposing the motion during its entire up and down path. How much work was done by air resistance during the entire path up and down.
Wfric= -f X
(-30)(62)= -1860J
is this correct
How did you get the distance X? Why isn't f = 25 N?
Equal amounts of work are done against friction going up and coming down.
You need to compute how far the ball goes up. Then multiply that by 2*f.
Its deceleration rate going up is
a = (Mg + f)/M = g + f/M = 14.8 m/s^2
It travels up for
T = (30 m/s)/a = 2.03 s
The distance travelled up is
(Vo/2)*T = 30.4 m
Wfriction = -2*30.4*25 = -1520 J