Question

A 75.0-kg cross-country skier is climbing a 3.0° slope at a constant speed of 2.00 m/s and encounters air resistance of 25.0 N.

Answer 1

Answer:

a) 12.01 W

b) 63.506 N

c) 11.82 s

Explanation:

Given data:

mass = 75 kg

slope of inclination = 3 degree

speed of skier is 2.00 m/s

air Resistance = 25 N

a) From the given information we have following relation

$F_n = mgsin\theta + F_a$

$F_n = 75\times 9.81 sin 3 + 25 = 63.506 N$

Power output

$P =F_n \times v$

$P = 63.506 \times 2 = 127.01 W$

b) average force can be calculate as

$F_n = mgsin\theta + F_a$

$F_n = 75 \times 9.81 sin 3 + 25$

F_n = 63.506 s

c)

force of acceleration is f = ma

solving for acceleration

$a = \frac{F}{m}$

$a = \frac{63.506}{75}$

a = 0.846 m/s^2

we know acceleration is given as $a =\frac{v -u}{t}$

solving for time t

$t = \frac{v-u}{a}$

$t =\frac{10-0}{0.846}$

t = 11.82 s