Ask question

Ask question

All categories

3 years ago

6

A person pulls a crate of mass M = 63 kg a distance 40.0 m along a horizontal floor by a constant force FP = 130 N, which acts a

t an angle of 23o from the horizontal. The floor is rough and has a coefficient of kinetic equal to 0
a) (2 pts.) Model: How will you model this problem?
b) (5 points) Visualize: Draw a free body diagram showing all forces acting on the crate. Be sure to label the direction of your positive unit vectors c) (5 points)
c) Visualize: Draw a pictorial representation of the crate and label all known quantities
d) (5 pts.) Solve: Calculate the work done by the gravitational force e) (15 pts.)
e) Solve: Calculate the normal force f (5 pts.)
f) Solve: Calculate the work done by the normal force g (10 pts.)
g) Solve: Calculate the magnitude of the frictional force h) (15 pts.)
h) Solve: Calculate the work done by the frictional force. i) (15 pts.)
i) Solve: Calculate the work done by the force exerted by the j) (3 pts.)
j) Solve: Calculate the net work done on the crate. _ person

1 answer:

GrogVix [38]3 years ago

7 0

Answer:

Check attachment for solution and diagrams

Explanation:

Given that,

Mass of crate m=63kg

Distance travelled d=40m

Horizontal force Fx=130N

Angle the force applied on cord makes with horizontal is θ=23°.

The weight of the crate is given by

W=mg

W=63×9.81

W=618.03N

Horizontal force Fx=130N

Resolving the applied Force F to the horizontal will give

Fx=FCos θ

F=Fx/Cos θ

F=130/Cos23

F=141.2N

a. Check attachment for model diagram

b. Check attachment for free body diagram

c. Check attachment for pictorial representation

d. Work done by gravitational force.

We, know that the body did not move upward, then the distance d=0

Work done is given as

W=F×d

So, d=0

W=F×0

W=0J

So, no work is done by gravity

e. Normal force?

Using newton law of motion

ΣFy = may

Since the body is not moving upward, then ay=0m/s²

N+141.2Sin23-618.03=0

N=618.03-141.2Sin23

N=562.86N

f. Work done by normal force.

The body is not moving upward, then the distance is zero

d=0

Work done by normal=normal force × distance

Wn=562.86×0

Wn=0J

No work is done by the normal force

g. Frictional force?

Since the coefficient of kinetic friction is zero, then the surface is frictionless

So, no frictional force is acting on the body

Fictional force is given as

Fr=μk•N

Given that, μk=0

Fr=0×562.86

Fr=0N

d. Work done by frictional force?

Since the frictional force Is zero, then, no work is done by friction

W(friction ) = frictional force × d

Here, the body moved a distance of 40m

W(fr)=0×40

W(fr)=0J

No work is done by friction

I. Work done by exerted force

The horizontal component of the exerted force is 130N and the body traveled a distance of 40m

Then, work done is given as

Workdone=force ×distance

Work done=130×40

W=5200J

W=5.2KJ

h. Net workdone?

Since no work is lost by friction, then, the net work done is equal to the work done by the exerted force.

Went, = work done by force exerted - work done by friction

Wnet=5200-0

Wnet, =5200

Wnet=5.2KJ

You might be interested in

A biologist is researching antibodies, which are tiny natural proteins in the blood of animals, and she is using methods develop

Sedbober [7]

Answer:

Her friend disagrees because the biologist is not working with nanotechnology which would enable her to be a nanotechnologist.

The biologist is only working with nano antibodies.

8 0

3 years ago

Read 2 more answers

In a cyclotron, the orbital radius of protons with energy 300 keV is 16.0 cm . You are redesigning the cyclotron to be used inst

Archy [21]

Answer:

16 cm

Explanation:

For protons:

Energy, E = 300 keV

radius of orbit, r1 = 16 cm

the relation for the energy and velocity is given by

$E = \frac{1}{2}mv^{2}$

So, $v = \sqrt{\frac{2E}{m}}$ .... (1)

Now,

$r = \frac{mv}{Bq}$

Substitute the value of v from equation (1), we get

$r = \frac{\sqrt{2mE}}{Bq}$

Let the radius of the alpha particle is r2.

For proton

So, $r_{1} = \frac{\sqrt{2m_{1}E}}{Bq_{1}}$ ... (2)

Where, m1 is the mass of proton, q1 is the charge of proton

For alpha particle

So, $r_{2} = \frac{\sqrt{2m_{2}E}}{Bq_{2}}$ ... (3)

Where, m2 is the mass of alpha particle, q2 is the charge of alpha particle

Divide equation (2) by equation (3), we get

$\frac{r_{1}}{r_{2}}=\frac{q_{2}}{q_{1}}\times \sqrt{\frac{m_{1}}{m_{2}}}$

q1 = q

q2 = 2q

m1 = m

m2 = 4m

By substituting the values

$\frac{r_{1}}{r_{2}}=\frac{2q}}{q}}\times \sqrt{\frac{m}}{4m}}=1$

So, r2 = r1 = 16 cm

Thus, the radius of the alpha particle is 16 cm.

8 0

3 years ago

Read 2 more answers

When the voltage is

gizmo_the_mogwai [7]

Explanation:

ask to nirajan shrestha

search on fb

he is good science teacher❣❣❣

7 0

2 years ago

Two objects that are not initially in thermal equilibrium are placed in close contact. After a while, the temperature of the cod

Dima020 [189]

Answer:

If the temperature of the colder object rises by the same amount as the temperature of the hotter object drops, then <u>the specific heats of both objects will be equal.</u>

Explanation:

If the temperature of the colder object rises by the same amount as the temperature of the hotter object drops when the two<u> objects of same mass</u> are brought into contact, then their specific heat capacity is equal.

<u>We can prove this by the equation of heat for the two bodies:</u>

<em>According to given condition,</em>

$\Delta T_1=\Delta T_2$

$\frac{Q_1}{m_1.c_1} = \frac{Q_2}{m_2.c_2}$

<em>when there is no heat loss from the system of two bodies then </em> $Q_1=Q_2$

$\frac{1}{m.c_1} =\frac{1}{m.c_2}$

$\Rightarrow c_1=c_2$

Thermal conductivity is ultimately affects the rate of heat transfer, however the bodies will attain their final temperature based upon their mass and their specific heat capacities.

The temperature of the colder object will rise twice as much as the temperature of the hotter object only in two cases:

when the specific heat of the colder object is half the specific heat of the hotter object while mass is equal for both.

OR

the mass of colder object is half the mass of the hotter object while their specific heat is same.

3 0

3 years ago

Derive a relation between Universal Gravitational Constant (G) and Acceleration due to gravity(g)?

snow_tiger [21]

Suppose earth is a soid sphere which will attract the body towards its centre.So, acc. to law of gravitation force on the body will be,
F=G*m1m2/R^2
but we now that F=ma
and here accleration(a)=accleration due to gravity(g),so
force applied by earth on will also be mg
replace above F in formula by mg and solve,
F=G*mE*m/R^2 ( here mE is mass of earth and m is mass of body)
mg=G*mEm/R^2
so,
g =G*mE/R^2

4 0

3 years ago

Read 2 more answers