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dangina [55]
3 years ago
9

Lien uses a spring scale to pull a block toward the right across the lab table. The scale reads 8 N. Which force should Lien con

clude is also 8 N?
Physics
1 answer:
lana66690 [7]3 years ago
6 0

Answer:

<h2>Frictional force</h2>

Explanation:

For the block placed on the table, there are several force acting on the body along the horizontal and vertical axis. All this forces tends to keep the body in a state of equilibrium.The forces acting along the horizontal are the moving force (Fm) and the frictional force (Ff).

Frictional force are forces that acts opposite to the force that causes the body to move (moving force).

If Lien uses a spring scale to pull a block toward the right across the lab table and the scale reads 8 N, this means that the force that causes the body to move is the 8N force (moving force).

Taking the sum of force along the horizontal;

\sum fx = ma_x

Since the body is static, max = 0

\sum fx = 0\\fm+(-Ff)= 0

Note that the frictional force acts is the force of opposition acting in the negative x direction.

fm = 0+Ff\\fm = Ff

Since Fm = 8N, Ff will also be equal to 8N.

Based on the above proof, Lien can also conclude that 8 N is a frictional force

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Answer:

F = 385.56 N

Explanation:

Given that,

Mass, m = 37.8 kg

He applies a horizontal force and cross the wooden floor.

We need to find the force that must be applied to move a dog with a constant speed of 1 m/s.

The coefficient of kinetic friction between the dog in the floor is 1.02.

The net force acting on it is given by :

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aev [14]

Heya!!

For calculate aceleration, lets applicate second law of Newton:

                                                    \boxed{F=ma}

                                                 <u>Δ   Being   Δ</u>

                                             F = Force = 183 N

                                            m = Mass = 367 kg

                                             a = Aceleration = ?

⇒ Let's replace according the formula and clear "a":

\boxed{a=183 \ N / 367\ kg}

⇒ Resolving

\boxed{ a = 0.49 \ m/s^{2}}

Result:

The aceleration is <u>0,49 meters per second squared (m/s²)</u>

Good Luck!!

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