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alexgriva [62]
3 years ago
9

A mass is suspended on a vertical spring. Initially, the mass is in equilibrium. Then, it is pulled downward and released. The m

ass then moves up and down between the "top" and the "bottom" positions. By definition, the period of such motion is the time interval it takes the mass to move:Mark all the correct statements among those provided below. a. from the top position to the bottom. b. from the equilibrium position to the bottom. c. from the bottom position to the top. d. from the equilibrium position to the bottom and then back to the equilibrium. e. from the equilibrium position to the top and then back to the equilibrium. f. from the equilibrium position to the top. g. from the top position to the bottom and then back to the top. h. from the bottom position to the top and then back to the bottom.
Physics
1 answer:
anastassius [24]3 years ago
8 0

Answer:

d. from the equilibrium position to the bottom and then back to the equilibrium.

g. from the top position to the bottom and then back to the top.

h. from the bottom position to the top and then back to the bottom.

Explanation:

It is the case of SHM or Simple Harmonic Motion. Firstly, there is a need to understand the time interval or time period. The standard definition of time period in simple harmonic motion is

"the time period required for the system to complete its one cycle"

Now one have to consider that  the system given above, the motion of mass attached to spring will follow the path of motion from equilibrium point to bottom to equilibrium point to top, then equilibrium point to the bottom and so on.

to choose right answer you must have to consider the option, in which the starting point and ending point of the mass is same. If mass starts from top, the time it will take to reach on top again, will be defined as its time period and so in the case of bottom or equilibrium as starting point. Hence, "d", "g" and "h" are right answers.

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Explanation:

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Complete Question

A 95 kg clock initially at rest on a horizontal floor requires a 650 N horizontal force to set it in motion. After the clock is in motion, a horizontal force of 560 N keeps it moving with a constant velocity. Find the coefficient of static friction and the coefficient of kinetic friction.

Answer:

The value for static friction is \mu_s =  0.60

The value for static friction is \mu_k =  0.70

Explanation:

From the question we are told that

The mass of the clock is m  =  95 \  kg

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Generally the kinetic frictional force is equal to the second horizontal force

So

      F _k  =  m  *  g  *  \mu_k

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     \mu_k =  0.70

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