Question

A skier of mass 58.0 kg starts from rest at the top of a ski slope of height 70.0 m . Part APart complete If frictional forces do −1.04×104 J of work on her as she descends, how fast is she going at the bottom of the slope? Take free fall acceleration to be g = 9.80 m/s2 . 31.8 m/s Previous Answers Correct Part BPart complete Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is μk = 0.250. If the patch is of width 68.0 m and the average force of air resistance on the skier is 150 N , how fast is she going after crossing the patch? 18.1 m/s Previous Answers Answer Requested Part C After crossing the patch of soft snow, the skier hits a snowdrift and penetrates a distance 2.30 m into it before coming to a stop. What is the average force exerted on her by the snowdrift as it stops her?

Answer 1

Answer:

A) v = 31.8 m/s

B) v_f = 18.1 m/s

C) F_avg = 4130.7 N

Explanation:

A) From law of conservation of energy, we know that;

mgh = W_f + ½mv²

v is the speed at which she is going at the bottom of the slope.

Thus, making v the subject, we have;

v = √[2gh - (2W_f)/m)]

We are given;

h = 70 m

m = 58 kg

W_f = 1.04 × 10⁴ J = 10400 J

Thus;

v = √[(2 × 9.8 × 70) - (2 × 10400)/58)]

v = 31.8 m/s

B) Total force will be given by the formula;

F_t = F_k + F_r

Where;

F_k is force of kinetic friction = mg•μ_k

μ_k = 0.25

So, F_k = 58 × 9.8 × 0.25

F_k = 142.1 N

We are given force of air resistance(F_r) as 150 N

Thus;

F_t = 142.1 + 150

F_t = 292.1 N

Final velocity is gotten from the formula;

v_i² - v_f² = 2F_t•L/m

Thus;

v_f = √[v_i² - (2F_t•L/m)]

Where, v_i = 31.8 m/s, F_t = 292.1, m = 58 kg, L = 68 m

Thus;

v_f = √[31.8² - (2 × 292.1 × 68/58)]

v_f = 18.1 m/s

C) the average force exerted on her by the snowdrift as it stops her is given by the formula;

F_avg = m•v_f²/2l

F_avg = 58 × 18.1²/(2 × 2.3)

F_avg = 4130.7 N