A skier is pulled by a tow rope up a frictionless ski slope that makes an angle of 5.6° with the horizontal. The rope moves parallel to the slope with a constant speed of 1.5 m/s. The force of the rope does 930 J of work on the skier as the skier moves a distance of 6.9 m up the incline. (a) If the rope moved with a constant speed of 2.1 m/s, how much work would the force of the rope do on the skier as the skier moved a distance of 6.9 m up the incline? At what rate is the force of the rope doing work on the skier when the rope moves with a speed of (b) 1.5 m/s and (c) 2.1 m/s?

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Netta00 Netta00 · Answer 1 · 2019-10-24T11:18:40+02:00

Answer:

a)The work done by force on the rope is 930 J

b)P=202.7 W

c)P=230.04 W

Explanation:

Given that

θ= 5.6°

Vo= 1.5 m/s

W=930 J

d= 6.9 m

a)

If the rope moved with a constant speed then change in the kinetic energy will be zero.

W₁ + W₂= ΔKE

ΔKE = 0

W₁ = - W₂

Given that W₁ = 930 J

W₂ = - 930 J

The work done by force on the rope is 930 J

b)

t= d/v

t= 6.9 / 1.5 = 4.6 s

P=W/t

P= 930/ 4.6 =202.7 W

P=202.7 W

c)

t= d/v

t= 6.9 /2.1 = 3.2 s

P=W/t

P= 930/ 3.2 =230.04 W

P=230.04 W