A movie stuntman (mass 80.0kg) stands on a window ledge 5.0 mabove the floor. Grabbing a rope attached to a chandelier, heswings down to grapple with

Answer:
a) The entwined foes start to slide across the floor with a speed of approximately $6.62\text{m/s}$.
b) They slide approximately $194.68\text{m}$.
Explanation:
a) To find the initial velocity of the entwined foes when they start to slide across the floor, we can equate the initial potential energy to the final kinetic energy.
The initial potential energy is given by the height the stuntman descends:
$P{E}_{\text{initial}}=m_{\text{stuntman}}·g·h$
where:
$m_{\text{stuntman}}$ is the mass of the stuntman (80.0 kg),
$g$ is the acceleration due to gravity $(9.8m/s^2)$, and
$h$ is the height the stuntman descends (5.0 m).
The final kinetic energy is given by the sum of the kinetic energies of the stuntman and the villain:
$K{E}_{\text{final}}=\frac{1}{2}·m_{\text{stuntman}}·v^2+\frac{1}{2}·m_{\text{villain}}·v^2$
where:
$m_{\text{villain}}$ is the mass of the villain (70.0 kg), and
$v$ is the common final velocity of both the stuntman and the villain.
Since there is no external work done and no energy losses, we can equate the initial potential energy to the final kinetic energy:
$m_{\text{stuntman}}·g·h=(\frac{1}{2}·m_{\text{stuntman}}+\frac{1}{2}·m_{\text{villain}})·v^2$
Now we can solve for $v$:
$v=\sqrt{\frac{2·m_{\text{stuntman}}·g·h}{m_{\text{stuntman}}+m_{\text{villain}}}}$
Substituting the given values:
$v=\sqrt{\frac{2·80.0\text{kg}·9.8{\text{m/s}}^2·5.0\text{m}}{80.0\text{kg}+70.0\text{kg}}}$
Simplifying the expression:
$v\approx 6.62\text{m/s}$
Therefore, the entwined foes start to slide across the floor with a speed of approximately 6.62 m/s.
b) To find the distance they slide, we need to consider the force of kinetic friction acting on the entwined foes. The force of kinetic friction is given by:
$F_{\text{friction}}=\mu ·F_{\text{normal}}$
where:
$\mu$ is the coefficient of kinetic friction (0.250), and
$F_{\text{normal}}$ is the normal force.
The normal force is equal to the weight of the system, which is the sum of the weights of the stuntman and the villain:
$F_{\text{normal}}=(m_{\text{stuntman}}+m_{\text{villain}})·g$
The work done by the force of kinetic friction is equal to the force of friction multiplied by the
distance:
$W_{\text{friction}}=F_{\text{friction}}·d$
Since there is no change in the kinetic energy of the entwined foes (the final kinetic energy is zero as they come to rest), the work done by the force of kinetic friction must be equal to the initial kinetic energy of the system:
$W_{\text{friction}}=K{E}_{\text{initial}}=\frac{1}{2}·(m_{\text{stuntman}}+m_{\text{villain}})·v^2$
Equating these two expressions and solving for $d$, we get:
$F_{\text{friction}}·d=\frac{1}{2}·(m_{\text{stuntman}}+m_{\text{villain}})·v^2$
$d=\frac{\frac{1}{2}·(m_{\text{stuntman}}+m_{\text{villain}})·v^2}{F_{\text{friction}}}$
Substituting the given values:
$d=\frac{\frac{1}{2}·(80.0\text{kg}+70.0\text{kg})·(6.62\text{m/s}{)}^2}{0.250}$
Simplifying the expression:
$d\approx 194.68\text{m}$
Therefore, the entwined foes slide approximately 194.68 meters across the floor.

Step 1:
a) First, let's find the initial potential energy of the stuntman when he is standing on the window ledge. The formula for potential energy is given by:
$PE=mgh$
where $m$ is the mass (80.0 kg), $g$ is the acceleration due to gravity (9.8 m/s²), and $h$ is the height (5.0 m).
Substituting the values, we have:
$PE=(80.0\text{kg})(9.8{\text{m/s}}^2)(5.0\text{m})$
Now, when the stuntman swings down and releases the rope, all his potential energy is converted into kinetic energy. The formula for kinetic energy is given by:
$KE=\frac{1}{2}m{v}^2$
where $m$ is the mass (80.0 kg) and $v$ is the velocity.
Setting the potential energy equal to the kinetic energy, we have:
$mgh=\frac{1}{2}m{v}^2$
Canceling out the mass, we can solve for the velocity:
$v=\sqrt{2gh}$
Substituting the given values, we get:
$v=\sqrt{2(9.8{\text{m/s}}^2)(5.0\text{m})}$
Simplifying this expression will give us the final answer for part (a).
Step 2:
b) To determine how far the entwined foes slide, we need to consider the friction force acting on them. The friction force can be calculated using the formula:
$f_{\text{friction}}=\mu N$
where $\mu$ is the coefficient of kinetic friction (0.250) and $N$ is the normal force.
The normal force is equal to the weight of the combined masses of the stuntman and the villain. The formula for weight is given by:
$W=mg$
where $m$ is the combined mass (80.0 kg + 70.0 kg) and $g$ is the acceleration due to gravity (9.8 m/s²).
The distance the entwined foes slide can be determined using the equation of motion:
$d=\frac{1}{2}a{t}^2$
where $d$ is the distance, $a$ is the acceleration (caused by friction), and $t$ is the time.
The time can be calculated using the formula:
$t=\frac{v}{a}$
where $v$ is the velocity and $a$ is the acceleration.
Substituting the values and solving these equations will give us the final answer for part (b).