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ZanzabumX [31]
3 years ago
7

Choose the 200 kg refrigerator. set the applied force to 400 n (to the right). be sure friction is turned off. what is the net f

orce acting on the refrigerator?

Physics
2 answers:
Firlakuza [10]3 years ago
5 0

The net force acting on the refrigerator is 400 N to the right.

<h3></h3><h3>FURTHER EXPLANATION</h3>

The net force or resultant force is the sum of all the forces acting on a body or an object in x and y axes.

  • Forces along the y-axis The forces that usually act on an object vertically (in the y-axis) are: gravitational force which is a downward force and the normal force which is an upward (perpendicular) force exerted by a surface on an object resting above it that keeps the object from falling.
  • Forces along the x-axis These include the force or forces applied to cause a left or right motion of an object along the horizontal plane (called the Applied Force) and the force that opposes the motion or friction.

In this problem the forces acting on the x and y - axes can be determined:

Along the x-axis:

  • gravitational force = -1960 N
  • normal force = +1960 N
  • Net force = -1960 N + 1960 N = 0

The gravitational force is the weight of the object obtained by multiplying the mass of the object (in kg) with the acceleration due to gravity, 9.8 m/s^2. It is given a negative (-) sign to indicate that it is a downward force.

Since the object is not falling through the surface, it can be assumed that the gravitational force and normal force are balanced. Hence, the size of the normal force is the same as the gravitational force but with the opposite direction indicated by the + sign for an upward force.

The forces along the x-axis are balanced  (i.e. net force is zero) so the object neither moves upward or downward.

Along the y-axis

  • applied force = +400 N
  • friction = 0
  • Net force = +400 N + 0 = +400 N

The applied force is +400 N. It is given a + sign to indicate that its direction is to the right.

The friction, as mentioned in the problem, is set to zero or "turned off".

The net force along the y-axis is +400. The forces are unbalanced so the object will move to the right as force is applied to it.

<h3>LEARN MORE</h3>
  • Balanced Forces brainly.com/question/760473
  • Unbalanced Forces brainly.com/question/4289010
  • Free body Diagrams brainly.com/question/12345810

Keywords: net force, resultant force

aleksandr82 [10.1K]3 years ago
3 0

The net force acting on the refrigerator is \fbox{400\,{\text{N}}}.

Further Explanation:

The net force acting on a body is the sum of all the force that are acted on the body considering the direction of action of the force. In other words, the net force acting on a body is the vector sum of all the forces acting on the body.

Given:

The mass of the refrigerator is 200\,{\text{kg}}.

The force acting on the refrigerator is 400\,{\text{N}}.

The friction on the surface is turned off.

Concept:

The refrigerator is kept on the floor and the friction on the floor is turned off. It means that there is no force opposing the motion of the refrigerator on the floor.

The different forces acting on the refrigerator areas shown in the figure attached.

There are the forces acting on the refrigerator in the vertical as well as in the horizontal direction. So, the net force acting on the refrigerator will be in horizontal as well as in the vertical direction.

<u>Net force in vertical direction: </u>

The forces acting on the refrigerator in the vertical direction are the weight of refrigerator acting vertically downward and the normal reaction on the refrigerator acting in the vertically upward direction.

The weight of the refrigerator is given by:

\fbox{\bf\ W = mg}

Here, W is the weight of the refrigerator, m is the mass of the refrigerator and g is the acceleration due to gravity.

Consider the value of Acceleration due to gravity to be 9.8\,{{\text{ m}}\mathord{\left/{\vphantom{{\text{m}}{{{\text{s}}^{\text{2}}}}}} \right.\kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}.

Substitute 200\,{\text{kg}} for m and 9.8\,{{\text{ m}}\mathord{\left/{\vphantom{{\text{m}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}} for g in above expression.

\begin{aligned}W&=\left({200\,{\text{kg}}}\right)\left({9.8\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}}\right)\\&=1960\,{\text{N}}\\\end{aligned}

The normal reaction on the refrigerator is the reaction force of the ground acting on the refrigerator. It is equal to the amount of force applied by the refrigerator on the ground. Therefore, the normal reaction acting on the refrigerator in the vertically upward direction is equal to the weight of the refrigerator but acting in opposite direction.

The net force on the refrigerator acting in the vertical direction is:

{F_v} = W - N

Here, {F_v} is the net force acting in vertical direction and N is the normal reaction on the refrigerator.

Substitute the value of W and N in above expression.

\begin{aligned}{F_v}&=1960\,{\text{N}}-1960\,{\text{N}}\\&=0\,{\text{N}}\\\end{aligned}

So, the net force acting on the refrigerator in vertical direction is 0 N.

<u>Net force in horizontal direction: </u>

Since the friction on the surface is turned off, there will be no opposing force to stop the motion of the refrigerator. The opposing force of friction acting on the refrigerator will be zero.

Friction force f=0\,{\text{N}}

The force acting on the refrigerator is acting to make the refrigerator move. The applied force on the refrigerator is 400\,{\text{N}}.

Applied force F=400\,{\text{N}}

The net horizontal force acting on the refrigerator is:

\begin{aligned}{F_h}&=400\,{\text{N}}-0\,{\text{N}}\\&=4{\text{00}}\,{\text{N}}\\\end{aligned}

Therefore, there is no net force on the refrigerator in the vertical direction and the net force on the refrigerator in the horizontal direction is \fbox{400\,{\text{N}}}.

Learn More:

1.  Acceleration of a box under friction brainly.com/question/7031524

2.  A ball falling under the acceleration due to gravity brainly.com/question/10934170

3. Stress developed in a wire brainly.com/question/12985068

Answer Details:

Grade: High School

Subject: Physics

Chapter: Force

Keywords:

Force, horizontal, vertical, weight, normal reaction, direction, net force, refrigerator, 400 N, 400N, applied force, normal, friction, gravity, acceleration.

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Answer:

a

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b

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    The resultant displacement in vertical direction (i.e the altitude change )

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Explanation:

From the question we are told that  

  The displacement in the morning is  S_{morning} =  (2200 \m , east) + (4000\ m\ north) + (100 \ m ,\ vertical)

   The displacement in the afternoon is  S _{afternoon}= (1300\ m ,\ west) + (2500 \ m ,\ north) - (300\ m ,\ vertical)

Generally the direction west is negative , the direction east is positive

                 the direction south is negative , the direction north is  positive

resultant displacement  is mathematically evaluated as  

    (2200 \m , east) +( - 1300\ m ,\ west) = 900 \ m \ east

     (4000\ m\ north)  + (2500 \ m ,\ north) = 6500  \ m ,\ north

     (100 \ m ,\ vertical) - (300\ m ,\ vertical) = -200 \ m

From the above calculation we see that at the end of the hiking the hiker (you) is  200 m below his/her(your) initial position

Generally from Pythagoras theorem , the resultant displacement in the north east direction is

      a  =  \sqrt{900^2 + 6500^2}

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Generally from Pythagoras theorem , the resultant displacement in vertical direction (i.e the altitude change )

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3 years ago
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DiKsa [7]

Answer:

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b). am = 0.56 m/s^2 (block),  aM = 0.28 m/s^2 (bucket)

Explanation:

a). We got  N = mg cos θ,

                  f = $\mu_s N$

                    = $\mu_s mg \cos \theta$

If the block is ready to slide,

T = mg sin θ + f

T = mg sin θ + $\mu_s mg \cos \theta$   .....(i)

2T = Mg ..........(ii)

Putting (ii) in (i), we get

$\frac{Mg}{2}=mg \sin \theta + \mu_s mg \sin \theta$

$M=2(m \sin \theta + \mu_s mg \cos \theta)$

$M=2 \times 10 \times (\sin 30^\circ+0.6 \cos 30^\circ)$

M = 20.392 kg

b). $(h-x_m)+(h-x_M)+(h'+x_M)=l$  .............(iii)

   Here, l = total string length

Differentiating equation (iii) double time w.r.t t, l, h and h' are constants, so

$-\ddot{x}-2\ddot x_M=0$

$\ddot x_M=\frac{\ddot x_m}{2}$

$a_M=\frac{a_m}{2}$   .....................(iv)

We got,   N = mg cos  θ

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∴ $T-(mg \sin \theta + f_K) = ma_m$

  $T-(mg \sin \theta + \mu_K mg \cos \theta) = ma_m$  ................(v)

Mg - 2T = Ma_M

$Mg-Ma_M=2T$

$Mg-\frac{Ma_M}{2} = 2T$    (from equation (iv))

$\frac{Mg}{2}-\frac{Ma_M}{4}=T$   .....................(vi)

Putting (vi) in equation (v),

$\frac{Mg}{2}-\frac{Ma_M}{4}-mf \sin \theta-\mu_K mg \cos \theta = ma_m$

$\frac{g\left[\frac{M}{2}-m \sin \theta-\mu_K m \cos \theta\right]}{(\frac{M}{4}+m)}=a_m$

$\frac{9.8\left[\frac{20.392}{2}-10(\sin 30+0.5 \cos 30)\right]}{(\frac{20.392}{4}+10)}=a_m$

$a_m= 0.56 \ m/s^2$

Using equation (iv), we get,

a_M= 0.28 \ m/s^2

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3 years ago
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