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nadezda [96]
3 years ago
7

Bob is pulling a 30kg filing cabinet with a force of 200n , but the filing cabinet refuses to move. the coefficient of static fr

iction between the filing cabinet and the floor is 0.80.
what is the magnitude of the friction force on the filing cabinet?
express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
puteri [66]3 years ago
8 0
The cabinet is being pulled with 200N and is being rested by a force equal to 200N. That is why it is not being moved. 


<span>Although the force of static friction can equal Fk=µs*F=m*g*µs=(30kg)*(9.8m/s^2)*(0.80)=235 N. It is not resisting the 200N force with 235N. Imagine if you pushed something with 200N and it pushed you back with 235N, especially a cabinet. You would think that the cabinet was alive.</span>
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Answer:

The electric field always decreases.

Explanation:

The electric field due to a point charge is given by :

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3 years ago
A box is placed on a conveyor belt that moves with a constant speed of 1.05 m/s. The coefficient of kinetic friction between the
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Answer:

The box stops in 0.139 seconds, after moving 7.29cm (0.0729m) backwards relative to the belt.

Explanation:

As the box is initially at rest relative to the earth, it is moving backwards with a speed of 1.05m/s relative to the belt. Then, the frictional force acts on the box to make it stop relative to the belt. So, we first have to write the equations of motion of the box in each axis:

x: f_k=ma\implies a=\frac{f_k}{m} \\\\y: N-mg=0\implies N=mg

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a=\frac{\mu_k mg}{m}=\mu_k g

Now, from the definition of acceleration we get:

a=\frac{v-v_0}{t}\implies t=\frac{v-v_0}{a}

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t=-\frac{v_0}{a}=-\frac{v_0}{\mu_k g}

(Don't worry about the negative sign. It will disappear because the initial velocity is also negative, since we take the box initially moving backwards)

Then, plugging in the given values, we calculate the time:

t=-\frac{(-1.05m/s)}{0.770(9.81m/s^{2})}=0.139s

In words, the time the box takes to stop sliding relative to the belt is 0.139s.

The displacement of the box in this time, is given by the kinematics formula:

v^{2}=v_0^{2}+2ax\implies x=-\frac{v_0^{2}}{2\mu_kg}

Finally, we calculate the displacement:

x=-\frac{(1.05m/s)^{2} }{2(0.770)(9.81m/s^2)}=-0.0729m=-7.29cm

This means that the box moves 7.29cm backwards relative to the belt.

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3 years ago
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Answer:

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3 years ago
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