At a playground, a 19kg child plays on a slide that drops through a height of 2.

Question

At a playground, a 19kg child plays on a slide that drops through a height of 2.3m. The child starts at rest at the top of the slide. On the way down, the slide does a nonconservative work of -361 J on the child. What is the childs speed at the bottom of the slide?

Explanation / Answer

Potential energy is mgh, on Earth, that's (19)(9.81)(2.3) m/s^2. Minus the 361 from that and invert the kinetic energy formula (1/2)mv^2, because what you have as potential has been translated to kinetic no matter what the route (well, if it's a slide the child has to keep sliding, so the route sort of matters). (19)(9.81)(2.3) - 361 = (1/2)(19)(v^2) =>[(19)(9.81)(2.3) - 361](2) / 19 = v^2 => v = sqrt of all that transferred messiness :) => v =~ 2.67 m/s Edit... oops, forgot to add that if you are under strict significant figure answers, it would be 2.7 m/s Edit again,.. sorry, put the units at the top wrong; it's in Newtons. I meant to place the m/s^2 right after the 9.81. Hope that didn't screw you up. Also, in real life, a slide doesn't end at ground level, but, oh well, I'm sure that's not the point.

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