Question

A 30-N stone is dropped from a height of 10 m and strikes the ground with a speed of 13 m/s.

Answer 1

Answer:

4.1 N

Step-by-step explanation:

We can solve this problem by using considerations about energy.

At the moment the stone is dropped, it has only gravitational potential energy:

$PE=mgh$

where

$mg=30 N$ is the weight of the stone

h = 10 m is the initial height of the stone

As the stone falls, part of this energy is converted into kinetic energy, while part into thermal energy due to the presence of the air friction, acting opposite to the motion of the stone:

$KE+E_t = \frac{1}{2}mv^2 + E_t$

where:

$m=\frac{mg}{g}=\frac{30}{9.8}=3.06 kg$ is the mass

v = 13 m/s is the final speed of the stone

$E_t$ is the thermal energy

The thermal energy is actually equal to the work done by the air friction on the stone:

$E_t=W=Fh$

where

F is the average force of friction

h = 10 m

Since the total energy must be conserved, we can combine the three equations, so we find:

$mgh=\frac{1}{2}mv^2+Fh$

And solving for F, we find the average force of air friction:

$F=\frac{mgh-0.5mv^2}{h}=\frac{(30)(10)-0.5(3.06)(13)^2}{10}=4.1 N$