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2 years ago

8

30. A box with mass 20kg is on a cement floor. The coefficient of static friction between the box and floor is 0.25. A man is pu

shing the box with a horizontal force of 35N. What is the magnitude of the force of static friction between the box and floor

1 answer:

Mnenie [13.5K]2 years ago

7 0

Answer:

84.05

Explanation:

F=mg×0.25
F=20x9.81×0.25
f=49.05N
F=35N

F=f+F

F=49.05+35

=84.05

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Answer: Option B

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Here we will call:

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2. $E_2$ : The energy after the mass is liberated by the spring

3. $E_3$ : The energy before the second string catch the mass

4. $E_4$ : The energy when the second sping compressed

so, the law of the conservations of energy says that:

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where $W_f$ is the work of the friction.

1. equation 1 is equal to:

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Solving for velocity, we get:

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2. Now, equation 2 is equal to:

$\frac{1}{2}MV_2^2-\frac{1}{2}MV_3^2 = U_kNd$

where M is the mass, $V_2$ the velocity in the situation 2, $V_3$ is the velocity in the situation 3, $U_k$ is the coefficient of the friction, N the normal force and d the distance, so:

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Volving for $V_3$ , we get:

$V_3 = 65.27 m/s$

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$\frac{1}{2}MV_3^2 = \frac{1}{2}K_2X_2^2$

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