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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:

a. t = \frac{v_{0}  +/- \sqrt{v_{0} ^{2} - gD} }{g}  b. D = v₀²/2g

Explanation:

Here is the complete question

A ball is thrown straight up from the ground with speed v₀ . At the same instant, a second ball is dropped from rest from a height D , directly above the point where the first ball was thrown upward. There is no air resistance

Find the time at which the two balls collide.

Express your answer in terms of the variables D ,v₀ , and appropriate constants..

t = ?!

Part B

Find the value of D in terms of v₀ and g so that at the instant when the balls collide, the first ball is at the highest point of its motion.

Express your answer in terms of the variables v₀ and g .

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Solution

The distance moved by the ball dropped from distance,D with velocity v₀, H₁ = D - (v₀t - gt²/2) = D + v₀t + gt²/2.

The distance moved by the ball thrown straight upward with velocity v₀ is H₂ = v₀t - gt²/2.

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Collecting like terms

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Using the quadratic formula,

t = \frac{-(-2v_{0} ) +/- \sqrt{(-2v_{0} )^{2} - 4 X g XD} }{2g} = \frac{2v_{0}  +/- \sqrt{4v_{0} ^{2} - 4gD} }{2g} = \frac{v_{0}  +/- \sqrt{v_{0} ^{2} - gD} }{g}

B. At its highest point, the velocity of the first ball, v = 0. Using v² = u² - 2gs where s = highest point of first ball when they collide and u = v₀.

0 = v₀² - 2gs

s = v₀²/2g.

Also, the time it takes the first ball to reach its highest point is gotten from v = u - gt. At highest point, v = 0 and u = v₀. So,

 0 = v₀ - gt₀

t₀ = v₀/g

Also H = s₁ + s where s₁  = distance moved by second ball in time t₀ for collision = v₀t₀ - gt₀²/2.

So, H = v₀t₀ - gt₀²/2 + v₀²/2g = v₀(v₀/g) - g(v₀/g)²/2 + v₀²/2g = v₀²/2g - v₀²/2g + v₀²/2g = v₀²/2g

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