Girl Climbing Rope: Avg Speed & Work Calculated

[Grey_117]

Homework Statement

A 55.0 kg student climbs a 4.00 m long rope and stops at the top.
(a) What must her average speed be in order to match the power output of a 200-W lightbulb?
(M/S)

(b) How much work does she do?
(J)

Homework Equations

Well if i had to guess...they might be

P=W/time total

1/2Mvf^2-1/2mvi^2

The Attempt at a Solution

 

[catkin]

It's OK to be unsure of the relevant equations but not OK when you only give the question. The forum is here to help you, not to do your homework for you.

So don't be shy; which equations do you even half-suspect may be relevant?

 

[Moetasim]

I would approach this problem by first converting the given information into SI units. The student's mass of 55.0 kg and the length of the rope of 4.00 m can be used to calculate the gravitational potential energy using the formula U=mgh, where g is the acceleration due to gravity (9.8 m/s^2). This gives us a value of 2156 J for the potential energy.

(a) To match the power output of a 200-W lightbulb, the student's average speed must be equal to the power output divided by the potential energy. This gives us an average speed of 0.093 m/s.

(b) The work done by the student can be calculated using the formula W=F*d, where F is the force applied and d is the distance moved. In this case, the force applied is equal to the weight of the student, which can be calculated using the formula F=mg. This gives us a force of 539 N. Multiplying this by the distance of 4.00 m, we get a work done of 2156 J, which is equal to the potential energy.

In conclusion, the student's average speed must be 0.093 m/s and the work done is 2156 J to match the power output of a 200-W lightbulb. It is important for scientists to use correct units and formulas in their calculations to ensure accurate results.