Question
Asked Oct 17, 2019
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A 0.400-kg particle has a speed of 1.50 m/s at point A and kinetic energy of 7.60 J at point B.

(a) What is its kinetic energy at A?

(b) What is its speed at B?

(c) What is the net work done on the particle by external forces as it moves from A to B?

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Expert Answer

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Step 1

Let the mass of the particle be denoted as m

Kinetic energy at point A and B be denoted as (K.E)A  and (K.E)B

The speed of the particle at point A and B be denoted as vA and vB .

 The kinetic energy of a particle at point is expressed as,

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1 K.E mv 2

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Step 2

For calculating the kinetic energy at the point A, (K.E)A  ,

Substitute 0.400 kg for m, 1.50 m/s for vA and solve for (K.E)A   as,

 

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(K.E)(0.400kg) (1.50 m/s) 0.45J

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Step 3

b)

 

For calculating the speed of the particle at the point B, use the expression for the kinetic energy (K.E)B  at the point B.

Substitute 7.60 J for (K.E...

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1 7.60J (0.400kg)(v) 15.2 B 0.4 6.16 m/s v

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