A 75.0-kg man steps off a platform 3.10 m above the ground. Hekeeps his legs straight as he falls, but at the moment his feettouch the ground his knee

Brittney Lord 2021-03-12 Answered
A 75.0-kg man steps off a platform 3.10 m above the ground. Hekeeps his legs straight as he falls, but at the moment his feettouch the ground his knees begin to bend, and, treated as aparticle, he moves an additional 0.60 m before coming torest.
a) what is the speed at the instant his feet touch theground?
b) treating him as a particle, what is his acceleration(magnitude and direction) as he slows down, if the acceleration isassumed to be constant?
c) draw his free-body diagram (see section 4.6). in termsof forces on the diagram, what is the net force on him? usenewton's laws and the results of part (b) to calculate the averageforce his feet exert on the ground while he slows down. expressthis force in newtons and also as a multiple of his weight.

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

Leonard Stokes
Answered 2021-03-14 Author has 16727 answers
Here are the detailed explanations:

image

Not exactly what you’re looking for?
Ask My Question
5
 
content_user
Answered 2021-09-30 Author has 11052 answers

a) The speed when his feet touch the ground is,

\(v=\sqrt{2gh}\)

\(=\sqrt{2(9.80\ m/s^2)(3.10m)}\)

\(=7.8\ m/s\)

b) Use kinematic equation to find the acceleration when he showdown is,

\(v^2=u^2+2as\)

since initial speed (u=0) is zero

\(v^2=2as\)

\(a=\frac{v^2}{2s}\)

\(=\frac{(7.80\ m/s^2)}{2(0.60\ m)}\)

\(=50.6\ m/s^2\)

c) Since the net force that his feet exert on the ground is same as the force that the ground exerts on his feet.

Apply newton second law for net force is,

\(F_{m}=ma\)

\(ma=F-mg\)

\(F=m(a+g)\)

\(=75.0\ kg)(50.6\ m/s^2+9.80\ m/s^2)\)

\(=4537.5\ N\)

Fraction of his weigh is,

\(\frac{F}{mg}=\frac{m(a+g)}{mg}\)

\(=\frac{a}{g}+1\)

\(=\frac{50.6\ m/s^2}{9.80\ m/s^2}+1\)

\(F=6.16mg\)

17

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Relevant Questions

asked 2021-03-26
A helicopter carrying dr. evil takes off with a constant upward acceleration of \(\displaystyle{5.0}\ \frac{{m}}{{s}^{{2}}}\). Secret agent austin powers jumps on just as the helicopter lifts off the ground. Afterthe two men struggle for 10.0 s, powers shuts off the engineand steps out of the helicopter. Assume that the helicopter is infree fall after its engine is shut off and ignore effects of airresistance.
a) What is the max height above ground reached by the helicopter?
b) Powers deploys a jet pack strapped on his back 7.0s after leaving the helicopter, and then he has a constant downward acceleration with magnitude 2.0 m/s2. how far is powers above the ground when the helicopter crashes into the ground.
asked 2021-02-27
An alpha particle (a He nucleus, containing two protons and two neutrons and having a mass of \(\displaystyle{6.64}\cdot{10}^{{-{27}}}\) kg) traveling horizontally at 35.6 km/s enters a uniform, vertical, 1.10 T magnetic field.
A) What is the diameter of the path followed by this alpha particle?
B) What effect does the magnetic field have on the speed of the particle?
C) What are the magnitude of the acceleration of the alpha particle while it is in the magnetic field?
D) What are the direction of the acceleration of the alpha particle while it is in the magnetic field?
asked 2021-03-24
A man with mass 70.0 kg stands on a platform withmass 25.0 kg. He pulls on the free end of a rope thatruns over a pulley on the ceiling and has its other end fastened tothe platform. The mass of the rope and the mass of the pulley canbe neglected, and the pulley is frictionless. The rope is verticalon either side of the pulley.
a) With what force does he have to pull so that he and theplatform have an upward acceleration of \(\displaystyle{1.80}\frac{{m}}{{s}^{{2}}}\)?
b) What is the acceleration of the rope relative to him?
asked 2021-03-31
A paraglider is flying horizontally at a constant speed.Assume that only two forces act on it in the vertical direction,its weight and a vertical lift force exerted on its wings by theair. The lift force has a magnitude of 1800 N.
(a) What is the magnitude and direction of the force that theparaglider exerts on the earth ?
(b)If the lift force should suddenly decrease to 1200 N, whatwould be the vertical acceleration of the glider ? For bothquestions, take the upward direction to be the + y direction.
asked 2021-03-17
The person weighs 170 lb. Each crutch makes an angle of 22.0 with the vertical. Half of the person's weight is supported by the cruches, the other half by the vertical forces exerted by the roundon his feet. Assuming that he is at rest and that the force exerted by the ground on the crutches acts along the crutches,determine
a) the smallest possible coefficient of friction between crutches and ground and
b) the magnitude of the compression force supported by each crutch.
asked 2021-03-24
A person bending forward to lift a load "with his back" rather than"with his knees" can be injured by large forces exerted on themuscles and vertebrae. A person is bending over to lift a200-N object. The spine and upper body are represented as auniform horizontal rod of weight 350 N, pivoted at the base of thespine. The erector spinalis muscle, attached at a pointtwo-thirds of the way up the spine, maintains the position of theback. The angle between the spine and this muscle is 12degrees. Find the tension in the back muscle and thecompressional force in the spine.
asked 2021-05-09
The dominant form of drag experienced by vehicles (bikes, cars,planes, etc.) at operating speeds is called form drag. Itincreases quadratically with velocity (essentially because theamount of air you run into increase with v and so does the amount of force you must exert on each small volume of air). Thus
\(\displaystyle{F}_{{{d}{r}{u}{g}}}={C}_{{d}}{A}{v}^{{2}}\)
where A is the cross-sectional area of the vehicle and \(\displaystyle{C}_{{d}}\) is called the coefficient of drag.
Part A:
Consider a vehicle moving with constant velocity \(\displaystyle\vec{{{v}}}\). Find the power dissipated by form drag.
Express your answer in terms of \(\displaystyle{C}_{{d}},{A},\) and speed v.
Part B:
A certain car has an engine that provides a maximum power \(\displaystyle{P}_{{0}}\). Suppose that the maximum speed of thee car, \(\displaystyle{v}_{{0}}\), is limited by a drag force proportional to the square of the speed (as in the previous part). The car engine is now modified, so that the new power \(\displaystyle{P}_{{1}}\) is 10 percent greater than the original power (\(\displaystyle{P}_{{1}}={110}\%{P}_{{0}}\)).
Assume the following:
The top speed is limited by air drag.
The magnitude of the force of air drag at these speeds is proportional to the square of the speed.
By what percentage, \(\displaystyle{\frac{{{v}_{{1}}-{v}_{{0}}}}{{{v}_{{0}}}}}\), is the top speed of the car increased?
Express the percent increase in top speed numerically to two significant figures.

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question
...