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

CoormaBak9

CoormaBak9

Answered question

2021-05-13

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 the movie's villian (mass 70.0 kg), whois standing directly under the chandelier.(assume that thestuntman's center of mass moves downward 5.0 m. He releasesthe rope just as he reaches the villian).
a) with what speed do the entwined foes start to slide acrossthe floor?
b) if the coefficient of kinetic friction of their bodies withthe floor is 0.250, how far do they slide?

Answer & Explanation

Adnaan Franks

Adnaan Franks

Skilled2021-05-15Added 92 answers

Given that
The mass of the stuntman is (ms)=80.0kg
The mass of the movies villian is (mv)=70.0kg
The distance between the window and floor is(y) =5.0m
The initial velocity is(u) =0
a)
By using the equation of motion we get that
The stuntman's speed before collision is (v0)=2gy
The speed after collision is given by the conservation law
v=(msms+mv)v0
b)
The coefficient of kinetic friction of their bodies with thefloor is (μk)=0.250
During sliding momentum is not conserved.So, from the work-energy theorem we get that
Here x is the distance moved
12mtv2=μkmtgh
x=v22μkg
Now substitute all the values you get the required answer
nick1337

nick1337

Expert2023-06-17Added 777 answers

Answer:
a) The entwined foes start to slide across the floor with a speed of approximately 6.62m/s.
b) They slide approximately 194.68m.
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:
PEinitial=mstuntman·g·h
where:
mstuntman is the mass of the stuntman (80.0 kg),
g is the acceleration due to gravity (9.8m/s2), 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:
KEfinal=12·mstuntman·v2+12·mvillain·v2
where:
mvillain 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:
mstuntman·g·h=(12·mstuntman+12·mvillain)·v2
Now we can solve for v:
v=2·mstuntman·g·hmstuntman+mvillain
Substituting the given values:
v=2·80.0kg·9.8m/s2·5.0m80.0kg+70.0kg
Simplifying the expression:
v6.62m/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:
Ffriction=μ·Fnormal
where:
μ is the coefficient of kinetic friction (0.250), and
Fnormal 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:
Fnormal=(mstuntman+mvillain)·g
The work done by the force of kinetic friction is equal to the force of friction multiplied by the
distance:
Wfriction=Ffriction·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:
Wfriction=KEinitial=12·(mstuntman+mvillain)·v2
Equating these two expressions and solving for d, we get:
Ffriction·d=12·(mstuntman+mvillain)·v2
d=12·(mstuntman+mvillain)·v2Ffriction
Substituting the given values:
d=12·(80.0kg+70.0kg)·(6.62m/s)20.250
Simplifying the expression:
d194.68m
Therefore, the entwined foes slide approximately 194.68 meters across the floor.
Vasquez

Vasquez

Expert2023-06-17Added 669 answers

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=(80.0kg)(9.8m/s2)(5.0m)
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=12mv2
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=12mv2
Canceling out the mass, we can solve for the velocity:
v=2gh
Substituting the given values, we get:
v=2(9.8m/s2)(5.0m)
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:
ffriction=μN
where μ 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=12at2
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=va
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).

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