1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
kotykmax [81]
2 years ago
12

A worker wants to load a 12 kg crate into a truck by sliding the crate up a straight ramp which is 2.5 m long and which makes an

angle of 30 degrees with the horizontal. The worker believes that he can get the crate to the very top of the ramp by launching it at 5 m/s at the bottom and letting go. But friction is not neglible; the crate slides 1.6 m upthe ramp, stops, and slides back down.
Required:
a. Assuming that the friction force actingon the crate is constant, find its magnitude.
b. How fast is teh crate moving when it reachesthe bottom of the ramp?
Physics
1 answer:
olga2289 [7]2 years ago
7 0

Answer:

a) The magnitude of the friction force is 55.851 newtons, b) The speed of the crate when it reaches the bottom of the ramp is 2.526 meters per second.

Explanation:

a) This situation can be modelled by the Principle of Energy Conservation and the Work-Energy Theorem, where friction represents the only non-conservative force exerting on the crate in motion. Let consider the bottom of the straight ramp as the zero point. The energy equation for the crate is:

U_{g,1}+K_{1} = U_{g,2}+K_{2}+ W_{fr}

Where:

U_{g,1}, U_{g,2} - Initial and final gravitational potential energy, measured in joules.

K_{1}, K_{2} - Initial and final translational kinetic energy, measured in joules.

W_{fr} - Work losses due to friction, measured in joules.

By applying the defintions of translational kinetic and gravitational potential energies and work, this expression is now expanded:

m\cdot g \cdot y_{1} + \frac{1}{2}\cdot m\cdot v_{1}^{2} =  m\cdot g \cdot y_{2} + \frac{1}{2}\cdot m\cdot v_{2}^{2} + \mu_{k}\cdot m \cdot g \cdot \cos \theta

Where:

m - Mass of the crate, measured in kilograms.

g - Gravitational acceleration, measured in meters per square second.

y_{1}, y_{2} - Initial and final height of the crate, measured in meters.

v_{1}, v_{2} - Initial and final speeds of the crate, measured in meters per second.

\mu_{k} - Kinetic coefficient of friction, dimensionless.

\theta - Ramp inclination, measured in sexagesimal degrees.

The equation is now simplified and the coefficient of friction is consequently cleared:

y_{1}-y_{2}+\frac{1}{2\cdot g}\cdot (v_{1}^{2}-v_{2}^{2}) = \mu_{k}\cdot \cos \theta

\mu_{k} = \frac{1}{\cos \theta} \cdot \left[y_{1}-y_{2}+\frac{1}{2\cdot g}\cdot (v_{1}^{2}-v_{2}^{2}) \right]

The final height of the crate is:

y_{2} = (1.6\,m)\cdot \sin 30^{\circ}

y_{2} = 0.8\,m

If \theta = 30^{\circ}, y_{1} = 0\,m, y_{2} = 0.8\,m, g = 9.807\,\frac{m}{s^{2}}, v_{1} = 5\,\frac{m}{s} and v_{2} = 0\,\frac{m}{s}, the coefficient of friction is:

\mu_{k} = \frac{1}{\cos 30^{\circ}}\cdot \left\{0\,m-0.8\,m+\frac{1}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)}\cdot \left[\left(5\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right] \right\}

\mu_{k} \approx 0.548

Then, the magnitude of the friction force is:

f =\mu_{k}\cdot m\cdot g \cdot \cos \theta

If \mu_{k} \approx 0.548, m = 12\,kg, g = 9.807\,\frac{m}{s^{2}} and \theta = 30^{\circ}, the magnitude of the force of friction is:

f = (0.548)\cdot (12\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot \cos 30^{\circ}

f = 55.851\,N

The magnitude of the force of friction is 55.851 newtons.

b) The energy equation of the situation is:

m\cdot g \cdot y_{1} + \frac{1}{2}\cdot m\cdot v_{1}^{2} =  m\cdot g \cdot y_{2} + \frac{1}{2}\cdot m\cdot v_{2}^{2} + \mu_{k}\cdot m \cdot g \cdot \cos \theta

y_{1}+\frac{1}{2\cdot g}\cdot v_{1}^{2} =y_{2} + \frac{1}{2\cdot g}\cdot v_{2}^{2} + \mu_{k}\cdot \cos \theta

Now, the final speed is cleared:

y_{1}-y_{2}+ \frac{1}{2\cdot g}\cdot v_{1}^{2} -\mu_{k}\cdot \cos \theta=  \frac{1}{2\cdot g}\cdot v_{2}^{2}

2\cdot g \cdot (y_{1}-y_{2}-\mu_{k}\cdot \cos \theta) + v_{1}^{2} = v_{2}^{2}

v_{2} = \sqrt{2\cdot g \cdot (y_{1}-y_{2}-\mu_{k}\cdot \cos \theta)+v_{1}^{2}}

Given that g = 9.807\,\frac{m}{s^{2}}, y_{1} = 0.8\,m, y_{2} = 0\,m, \mu_{k} \approx 0.548, \theta = 30^{\circ} and v_{1} = 0\,\frac{m}{s}, the speed of the crate at the bottom of the ramp is:

v_{2}=\sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot [0.8\,m-0\,m-(0.548)\cdot \cos 30^{\circ}]+\left(0\,\frac{m}{s} \right)^{2}}

v_{2}\approx 2.526\,\frac{m}{s}

The speed of the crate when it reaches the bottom of the ramp is 2.526 meters per second.

You might be interested in
Female scientist who came to america to study starts at harvard
krok68 [10]
Is that a question? If it is not what its the question?
7 0
3 years ago
Do we include signs if we calculate Electric field strength?
expeople1 [14]
Yes because if not people wouldn't understand how did you calculate electric field strength.
6 0
2 years ago
You are falling off the edge.what should you do to avoid falling..?
Nadya [2.5K]
Lean your shoulders back and your waist forwards. Use your arms as a counter weight.
3 0
3 years ago
Read 2 more answers
Does the planet exert a torque on the meteoroid with respect to the center of mass of the planet? Why or why not?
ra1l [238]

Answer:

yes

Explanation:

because the planet exerts a centripetal force on the meteorold

4 0
2 years ago
A car travels initially at 24 m/s, until it enters the highway. If the car accelerates at 4 m/s^2 for a 96 meters, what is the c
marishachu [46]
  • initial velocity=u=24m/s
  • Acceleration=a=4m/s^2
  • Distance=s=96m
  • Final velocity=v

Using 3rd equation of kinematics

\boxed{\Large{\sf v^2-u^2=2as}}

\\ \Large\sf\longmapsto v^2=u^2+2as

\\ \Large\sf\longmapsto v^2=24^2+2(4)(96)

\\ \Large\sf\longmapsto v^2=576+768

\\ \Large\sf\longmapsto v^2=1344

\\ \Large\sf\longmapsto v=\sqrt{1344}

\\ \Large\sf\longmapsto v=36.6m/s

3 0
3 years ago
Other questions:
  • Answer these questions fast please:-
    13·1 answer
  • What is a planet <br><br> Please answer this <br> ASAP
    8·2 answers
  • Why does an object in motion stay in motion unless acted on by an unbalanced force?
    7·2 answers
  • How many lymphatic organs are located in the abdominal cavity
    11·1 answer
  • A tennis ball is dropped from 1.3 m above
    10·1 answer
  • In our usual coordinate system( +x to the right, +y up (away from the center of the Earth), +z out of the page toward you), what
    15·1 answer
  • If a microwave oven is rated at 2400 watts and it is connected at a 120 volt
    13·2 answers
  • True or false in order for light to be emitted electrons must move from high energy state to a lower energy state?
    9·2 answers
  • The design of interior spaces is relatively unimportant to good<br> architecture?
    7·1 answer
  • Một điện lượng 60mC dịch chuyển qua tiết diện thẳng của dây dẫn trong khoảng
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!