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uranmaximum [27]
1 year ago
12

A ball on a string travels once around a circle with a circumference of 2. 0 m. The tension in the string is 5. 0 n.

Physics
1 answer:
Schach [20]1 year ago
4 0

The work done by the ball on a string is 0J if a ball is travelling around as circle with a circumference of 2.0m

We know that circumference or perimeter of circle=2πr where r is the radius of the circle.

So, we have circumference=2m

Therefore,2πr=2

=>r=2/2π

=>r=(1/π) m

Now, we know very well that when a body moves a displacement dx under the action of force, some amount of work is done by the force on the body and that force is given by

W=\int\limits {F}. \, dx

where F is defined as the force acting on the body

and dx is the displacement of the body.

On expanding the above formula, we get

=>W=Fdxcosθ where cosθ is the angle between the force and displacement of the body.

Since Tension is actually centripetal force which is acting along the center of the circle where ball displacement is in perpendicular direction to the tension.

It means angle between Tension and ball's displacement is 90°. So,cos90°=0

Therefore, W=Fdxcos90

=>W=0J.

Hence, amount of work done is 0J.

To know more about work, visit here:

brainly.com/question/18094932

#SPJ4

(Complete question) is:

A ball on a string travels once around a circle with a circumference of 2. 0 m. The tension in the string is 5. 0 N. What amount of work done by the ball on the string?

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Answer:

the force P required for impending motion is 132.3 N

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Explanation:

Given that:

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A free flow body diagram illustrating what the question represents is attached in the file below;

The given condition from the question let us realize that ; the casters are locked to prevent the tires from rotating.

Thus; considering the forces along the vertical axis ; we have :

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Considering the forces on the horizontal axis:

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\mu_s N_A + \mu_s N_B  = P

\mu_s( N_A +  N_B)  = P

replacing our value from equation (1)

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b) Since the horizontal distance between the casters A and B is 480 mm; Then half the distance = 480 mm/2 = 240 mm = 0.24 cm

the largest value of "h" allowed for  the cabinet is not to tip over is calculated by determining the limiting condition  of the unbalanced torque whose effect is canceled by the normal reaction at N_A and it is shifted to N_B:  

Then:

\sum M _B = 0

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