#### Answer

(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

#### Work Step by Step

(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.