# Minimum work needed to climb to the top of a peak

derpydashie6167
Homework Statement:
A 71.5-kg hiker starts at an elevation of 1300 m and climbs to the top of a peak 2680 m high. What is the minimum work required of the hiker?
Relevant Equations:
w=(Fcos[theta])d
F=ma
PE=mgh
f=ma=71.5*9.8 = 700.7 I know this is not right because he is also going up against gravity but I don't know what else to use for acceleration. I don't know the angle but I assume it is a 90 degree cliff.
w = (Fcos90)1380 = 0. But zero is not the correct answer.

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Kyouran
The angle of the cliff does not matter. The angle theta is also not the angle of the cliff, it is the angle between the force applied and the path taken, and is thus definitely not 90 degrees. The work done is then the path integral of the inner product between the force applied and the displacement. However, in this particular case you don't need to know any of those. Think about how work and energy are related.

Homework Helper
Gold Member
Homework Statement:: A 71.5-kg hiker starts at an elevation of 1300 m and climbs to the top of a peak 2680 m high. What is the minimum work required of the hiker?
Relevant Equations:: w=(Fcos[theta])d
F=ma
PE=mgh

f=ma=71.5*9.8 = 700.7 I know this is not right because he is also going up against gravity but I don't know what else to use for acceleration. I don't know the angle but I assume it is a 90 degree cliff.
w = (Fcos90)1380 = 0. But zero is not the correct answer.
In my opinion this is a poorly stated problem, but I understand what its author wants you to do. You need to find the minimum work done by the climber against gravity. Presumably that is minimum when the climber's kinetic energy does not change as (s)he climbs. Does that help? Think work-energy theorem.

Also, you seem to have a serious misconception about work. If the cliff is vertical, that means a 90o angle with respect to the horizontal. Angle θ in the expression for work is the angle between the force and the displacement vectors. Do you see why that angle is not 90o?