Question

A 0.017-kg acorn falls from a position in an oak tree that is 18.5 meters above the ground.

Answer 1

Answer:

Velocity= 19.05 m/s

Kinetic energy= 3.08 J

Explanation:

The corn possess potential energy. Feom law of conservation of energy, all the potential eneegy will be converted to kinetic energy, with negligible friction.

Potential energy, PE= kinetic energy, KE

Potential energy is given by mgh while kinetic energy is given by $0.5mv^{2}$

Equating

$mgh=0.5mv^{2}$

Making v the subject of the formula then

$v=\sqrt {2gh}$

Substituting 9.81 m/s2 for g and 18.5 m for h then the velocity will be

$v=\sqrt {2\times 9.81\times 18.5}=19.0517715711689\ m/s\\v\approx 19.05\ m/s$

To get the kinetic eneegy, since PE=KE AND KE is

$KE=0.5mv^{2}$

Substituting v with 19.05 and m with 0.017 kg we have

$0.5\times 0.017\times 19.05^{2}=3.08467125/ J\approx 3.08 J$

This can also be achieved by mgh since PE=KE hence 0.17*9.81*18.5=3.085245 J and the results are slightly different due to small energy losses and rounding off of values.