In a rock climbing gymnasium a climber with a wt = 125 lbs is climbing vertically for a distance of s = 25 ft with a constant velocity. it takes her t = 3 min to climb up the wall. a) what is the work done by the climber? (2 pts.) b) what is the power output of the climber during the ascent? (2 pts.) c) what is her potential energy after climbing the distance s?

a)  W = 3125 lb-ft , b)  P = 17.4 lb ft / s  and  c) U = 3125 lb ft

Explanation:

a) work is

     W = F. x

The bold is vector, we can work in a scalar way

    W = F x cos θ

As constant speed rises, the force applied must be equal to its weight, so that there is no acceleration

    F = W = mg

Work is

   W = mg x cos 0º

The pounds are units of weight

   W = 125 25

   W = 3125 lb-ft

b) The power ex work by time unit

    P = W / t

    P = 3125 / (3 60)

    P = 17.4 lb ft / s = 17.4 Hp

c) The equation for power energy is

    U = m g h

    U = 125 25

    U = 3125 lb ft

it's a b and

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y