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Nesterboy [21]
3 years ago
9

A filing cabinet weighing 556 N rests on the floor. The coefficient of static friction between it and the floor is 0.68, and the

coefficient of kinetic friction is 0.56. In four different attempts to move it, it is pushed with horizontal forces of magnitudes
(a) 222 N,
(b) 334 N,
(c) 445 N, and
(d) 556 N. For each attempt, calculate the magnitude of the frictional force on it from the floor. (The cabinet is initially at rest.) (e) In which of the attempts does the cabinet move?
Physics
1 answer:
zvonat [6]3 years ago
5 0

Explanation:

"Static friction is a force that keeps an object at rest. It must be overcome to start moving the object."

(556 x 0.68) = static friction of 378.08N. before movement occurs.

The  forces (a) and (b) will not move it.  

Each will incur a frictional force preventing movement equal to itself, = 222N. and 334N. respectively.

Forces (c) and (d) will move it, and accelerate it.

Forces (c) and (d) will both encounter friction of (556 x 0.56) = 311.36N. when the cabinet is moving.

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The velocity of a mass on a spring can be calculated by using the law of conservation of energy. In fact, the total energy of the mass-spring system is equal to the sum of the elastic potential energy (U) of the spring and the kinetic energy (K) of the mass:

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Since the total energy E must remain constant, we can notice the following:

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Based on these observations, we can conclude that the velocity of the mass is at its maximum value when the displacement is zero, so the correct option is A.


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A 2kg block has 70J of KE. It then travels 1.5 meters up a hill. As it travels up the hill friction does -12J of work on the blo
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Answer:

v = 5.34[m/s]

Explanation:

In order to solve this problem, we must use the theorem of work and energy conservation. This theorem tells us that the sum of the mechanical energy in the initial state plus the work on or performed by a body must be equal to the mechanical energy in the final state.

Mechanical energy is defined as the sum of energies, kinetic, potential, and elastic.

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In the initial state, we only have kinetic energy, potential energy is not had since the reference point is taken below 1.5[m], and the reference point is taken as potential energy equal to zero.

In the final state, you have kinetic energy and potential since the car has climbed 1.5[m] of the hill. Elastic energy is not available since there are no springs.

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Now we can use the first statement to get the first equation:

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70 - 12 = \frac{1}{2}*2*v^{2}+2*9.81*1.5

58 = v^{2} +29.43\\v^{2} =28.57\\v=\sqrt{28.57}\\v=5.34[m/s]

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