Question

A child, initially at rest at the top of 3.55 m high slide, slides down to a height of .480 m. At the bottom of the slide, the child has a speed of 2.45 m/s. If the mass of the child is 34.5 kg, what is the change in the mechanical energy of the child?

Answer 1

Answer on Question #43397, Physics, Mechanics | Kinematics | Dynamics

A child, initially at rest at the top of $3.55\mathrm{m}$ high slide, slides down to a height of $.480\mathrm{m}$. At the bottom of the slide, the child has a speed of $2.45\mathrm{m/s}$. If the mass of the child is $34.5\mathrm{kg}$, what is the change in the mechanical energy of the child?

Solution.

Mechanical energy is the sum of potential and kinetic energy:

$$E _ {m e c h} = E _ {p o t} + E _ {K}$$

At the top and at the bottom of the slide corresponding child's energy is:

$$E _ {1} = E _ {p o t 1} + E _ {K 1} = m g h _ {1} + 0 = m g h _ {1}$$ $$E _ {2} = E _ {p o t 2} + E _ {K 2} = m g h _ {2} + \frac {m V ^ {2}}{2}$$

The change in the mechanical energy:

$$\Delta E = E _ {1} - E _ {2} = m g (h _ {1} - h _ {2}) - \frac {m V ^ {2}}{2}$$

Numerically:

$$\Delta E = 34.5 k g \cdot 9.8 \frac {m}{s ^ {2}} \cdot (3.55 - 0.48) - \frac {34.5 k g \cdot \left(2.45 \frac {m}{s}\right) ^ {2}}{2} \approx 934 J$$

Answer: 934 J

http://www.AssignmentExpert.com/