Answer:
a) 101536 J
b) 84613 N
c) 210904 J
Step-by-step explanation:
a)Right before the impact, the pilot has a kinetic energy of:
![E_k = mv^2/2 = 82*50^2/2 = 102500 J](https://img.qammunity.org/2021/formulas/physics/high-school/ofqawqfar1nb9lsf5oa56srkrtxphptog8.png)
The snow did some work to slow his speed/kinetic energy down to 0. So his initial kinetic energy is converted to potential energy (of s = 1.2 m) and the work done by snow:
![E_k = E_p + W_s](https://img.qammunity.org/2021/formulas/physics/high-school/ynw04l7lihwnjfe0ccye80cm4e63ylt13r.png)
![103500 = mgs + W_s](https://img.qammunity.org/2021/formulas/physics/high-school/tgkbzfwy1kr6xeu1on6s5udzojxki3v8pv.png)
![W_s = 103500 - mgs = 103500 - 82*1.2*9.8 = 101536 J](https://img.qammunity.org/2021/formulas/physics/high-school/5iksy1vac03bpx0xfdugop4rp422uk19ix.png)
b) As work is the product of force over distance s = 1.2m, we can calculate the average force exerted on him as:
![F = W_s / s = 101536 / 1.2 = 84613 N](https://img.qammunity.org/2021/formulas/physics/high-school/izycsnynwpev7w3sej58g67ui8x9okpzm0.png)
c) His potential energy when jumping off from the aircraft at height h = 390 m is
E = mgh = 82*9.8*390 = 313404 J
As he reaches the snow, air resistance has done some work on him and reduces his mechanical energy to 102500 J, so that work must be:
313404 - 102500 = 210904 J