Question

A 40 kg child slides down a 5.0m long slide as shown. as the child slides, there is a kinetic friction force between the child and the slide with a coeffiecient of kinetic friction of 0.20. how much work (including the sign) did the gravitational force do to the child when the child slides from the top of the slide to the bottom?

Answer 1

The gravitational work done on the child, when sliding down the slide is positive 1960 Joules. This is calculated using the formula for gravitational work - W=mgd, and considering that the work is positive as the direction of force and displacement are the same.

Step-by-step explanation:

The gravitational work done on the child can be calculated using the formula Work=W=mgd Cos theta (where m is mass, g is gravity, d is distance/height and theta is angle). Given that the object's movement is vertical and involves gravity, the angle is 0-degree and Cos 0 equals to 1. Hence, the formula simplifies to W=mgd.

Substituting the given figures: m=40kg, g=approximately 9.8 m/s² (gravity), d=5m, we get W=40kg * 9.8 m/s² * 5m=P>

So, the work done by the gravitational force on the child sliding down the slide is 1960 Joules.

As the child is moving downward, the direction of the force (gravity) and the displacement are the same, so the work is positive. Therefore, the

direct answer in two lines

is: The gravitational work done on the child sliding down the slide is positive 1960 Joules.

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