# A 55.6 kg hiker climbs to the top of a hill whose shape can be approximated by the equation y=ax^2 , where y is the height in meters above the ground level. The value of a is 3.1 m^-2 , and both x...

A 55.6 kg hiker climbs to the top of a hill whose shape can be approximated by the equation y=ax^2 , where y is the height in meters above the ground level. The value of a is 3.1 m^-2 , and both x and y are equal to 0 m at the base of the hill. What is the total work done by gravity on the hiker after he has climbed from the bottom of the hill to a height y= 50.0 m?

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### 2 Answers

If you take the x axis horizontal and the y axis vertical for a total variation of the height of `Delta(y) =50 m` the work done by hiker to increase its potential energy is:

`W = m*g*Delta(y) = 55.6*9.81*50 =27271.8 J`

The work done by gravity is simply this work above with a sign minus in front of it (that show that the force of gravity is downwards and the higer is climbing upwards)

`W_g=-27271.8 J`

This happens because the gravitational force is a conservative force or in other words it does not depend on the path followed between initial and final positions.

You need to evaluate the work done by gravity, hence, you need to set up the following integral, such that:

`W = int_0^50 (m*g*a*x^2)dx`

`W = m*g*a*int_0^50 x^2 dx`

The problem provides the mass of hiker, m = 55.6 kg, the leading coefficient `a = 3.1 m^(-2)` and you may consider the gravitational acceleration `g = 9.8 m*s^(-2)` , such that:

`W = 55.6*9.8*3.1*(x^3/3)|_0^50`

`W = (55.6*9.8*3.1)/3(50^3 - 0^3)`

`W = 70380333 Kg*m^2/s^2`

**Hence, evaluating the work done by gravity,under the given conditions, yields `W = 70380333 Kg*m^2/s^2` .**

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