## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

(a) $F = 100~N$ (b) The power output at t = 2.0 s is 400 W The power output at t = 4.0 s is 800 W The power output at t = 6.0 s is 1200 W
(a) We can find the sprinter's acceleration as: $x = \frac{1}{2}at^2$ $a = \frac{2x}{t^2}$ $a = \frac{(2)(50~m)}{(7.0~s)^2}$ $a = 2.0~m/s^2$ We then find the horizontal force $F$ acting on the sprinter. $F = ma$ $F = (50~kg)(2.0~m/s^2)$ $F = 100~N$ (b) We can find the speed at t = 2.0 s $v = a~t$ $v = (2.0~m/s^2)(2.0~s)$ $v = 4.0~m/s$ We then find the power at t = 2.0 s $P = F~v$ $P = (100~N)(4.0~m/s^)$ $P = 400~W$ The power output at t = 2.0 s is 400 W. Next, we find the speed at t = 4.0 s: $v = a~t$ $v = (2.0~m/s^2)(4.0~s)$ $v = 8.0~m/s$ We can find the power at t = 4.0 s: $P = F~v$ $P = (100~N)(8.0~m/s^)$ $P = 800~W$ The power output at t = 4.0 s is 800 W. We can find the speed at t = 6.0 s $v = a~t$ $v = (2.0~m/s^2)(6.0~s)$ $v = 12.0~m/s$ We can find the power at t = 6.0 s $P = F~v$ $P = (100~N)(12.0~m/s^)$ $P = 1200~W$ The power output at t = 6.0 s is 1200 W.