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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valentin68's profile pic

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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.

sciencesolve's profile pic

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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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