Question #8a41c

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Question #8a41c

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Answer 1 · 2017-03-29T03:00:01

$15$ $15$ m/s

Explanation:

We first assume that no energy is lost while sliding down (i.e. no friction).

The gravitational potential energy at the top of the slide is given by $G P E = m g h = 74 k g \cdot 9.81 \frac{m}{s} \cdot 11.8 m = 8566.092$ $GPE=mgh=74kg*9.81m/s*11.8m=8566.092$ Joules. Since he starts from rest, kinetic energy is $0$ $0$ .

At the bottom of the slide, his gravitational potential energy is $0$ $0$ . However, he has gained kinetic energy equivalent to $K E = \frac{m v^{2}}{2} = \frac{74 J \cdot v^{2}}{2} = 37 v^{2}$ $KE=(mv^2)/2=(74J*v^2)/2=37v^2$ Joules.

Since we assumed that no energy is lost, the sum of the energy on top of the slide is equal to the sum of the energy at the bottom of the slide. Therefore, $8566.092 + 0 = 0 + 37 v^{2}$ $8566.092+0=0+37v^2$ . Since $v \geq 0$ $v≥0$ , solving it gives the velocity as $15.215650$ $15.215650$ m/s. Rounding to the appropriate number of significant places gives $15$ $15$ m/s.

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Question #8a41c

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