Answer
(a) $W = -9.0 \times 10^4 ~J$
(b) $F = 8.2\times 10^4 ~N$
(c) $W = -2.3 \times 10^5 ~J$
Work Step by Step
(a) The work done by the snow is equal to the change in energy. We need to consider $\Delta KE$ and $\Delta PE$ starting from the moment of impact with the snow until the person came to a stop. Therefore,
$W = \Delta KE + \Delta PE$
$W = -\frac{1}{2}mv^2 - mgd$
$W = \frac{1}{2}(88 ~kg)(45 ~m/s)^2 -(88 ~kg)(9.80 ~m/s^2)(1.1 ~m)$
$W = -9.0 \times 10^4 ~J$
(b) Let $F$ be the magnitude of the force. Then ($F \cdot d$) is equal to the magnitude of the work:
$F\cdot d = 9.0 \times 10^4 ~J$
$F = \frac{9.0 \times 10^4 ~J}{d} = \frac{9.0 \times 10^4 ~J}{1.1 ~m}$
$F = 8.2\times 10^4 ~N$
(c) We can use work and energy to solve this question.
$PE + W = KE$
$mgy - W = \frac{1}{2}mv^2 $
$W = \frac{1}{2}mv^2 - mgy$
$W = \frac{1}{2}(88 ~kg)(45 ~m/s)^2 - (88 ~kg)(9.80 ~m/s^2)(370 ~m)$
$W = -2.3 \times 10^5 ~J$