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A 3.00-kg ball swings rapidly in a complete vertical circle of radius 2.00 m by a light string that is fixed at one end. The ball moves so fast that the string is always taut and perpendicular to the velocity of the ball. As the ball swings from its lowest point to its highest point:

A. The work done on it by gravity is +118 J and the work done on it by the tension in the string is -118 J.
B. The work done on it by gravity is -118 J and the work done on it by the tension in the string is zero.
C. The work done on it by gravity and the work done on it by the tension in the string are both equal to zero.
D. The work done on it by gravity and the work done on it by the tension in the string are both equal to -118 J.
E. The work done on it by gravity is -118 J and the work done on it by the tension in the string is +118 J.

Answer :

Answer:

Option B

The work done on it by gravity is -118 J and the work done on it by the tension in the string is zero.

Explanation:

Total displacement, s is given by

s=r+r where r is the radius, which for this case is 2m

s=2+2=4m

Since work, W is a product of force and displacement

W=Fs but F=mg where F is force, s is displacement, m is mass and g is acceleration due to gravity, taken as [tex]9.81 m/s^{2}[/tex]

W=m(-g)s, here, g is negative because acceleration due to gravity is directed downwards

3Kg*-9.81*4=-117.72 J approximately -118 J

The work done due to tension is given by

[tex]W=Tscos\theta[/tex] where [tex]\theta[/tex] is the angle between displacement and force. Since tension and displacement are perpendicular, then the angle [tex]\theta[/tex] is [tex]90^{o}[/tex]hence

W=0 because cos 90 is zero

Therefore, the correct option is B

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Rewritten by : Barada

The work done on the ball by gravity as it swings from its lowest point to its highest point is -118 J.

The given parameters;

mass of the ball, m = 3 kg

radius of the circle, r = 2 m

The work done on the ball by gravity is the change in the mechanical energy of the ball.

W = ΔK.E + ΔP.E

at the maximum height, the final velocity = 0

W = ΔP.E

W = mg(h₂ - h₁)

where;

  • h₂ is the highest point of the vertical circle = 2r = 2 x 2 = 4 m
  • h₁ is the lowest point = 0

W = mg(4 - 0)

W = 4mg

W = 4 x (3) x (-9.8)

W = - 117.6 J

W = -118 J

Thus, the work done on the ball by gravity is -118 J.

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