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
