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A hand pushes two blocks, A and B, along a frictionless table for a distance d. When the hand starts to push, the blocks are moving with a speed of 2 m/s. Suppose that the work done on block A by the hand during a given displacement is 10 J. Determine the final energy of each block. Mass of A is 4kg, mass of B is 1kg.

Respuesta :

  net work = change in kinetic energy 

for Block B, we just have the force from block A acting on it 
F(ab)d= .5(1)vf² - .5(1)(2²) 
F(ab)d= .5vf² - 2 

Block A, we have the force from the hand going in one direction and the force of block B on A going the opposite direction 

10-F(ba)d = .5(4)vf² - .5(4)(2²) 
10-F(ba)d = 2vf² - 8 
F(ba)d = 18 - 2vf² 

now we have two equations: 
F(ba)d = 18 - 2vf² 
F(ab)d= .5vf² - 2 

since the magnitude of F(ba) and F(ab) is the same, substitute and find vf (I already took into account the direction when solving for F(ab) 

10-.5vf² + 2 = 2vf² - 8 
12 - .5vf² = 2vf² - 8 
20 = 2.5vf² 
vf² = 8 

they both will have the same velocity 
KE of block A= .5(4)(2.828²) = 16 J 
KE of block B=.5(1)(2.828²) = 4 J
ogorwyne

The final mass of block A = 15.99 joules

The final mass of Block B = 3.99 joules

Speed = 2ms

Displacement = 10j

mass of A = 4

Mass of B = 1

Change in kinetic energy

For B = 1

0.5Avf²-0.5AVS²

= 0.5*1*vf²-0.5*1*2²

= 0.5vf²-2 ---(1)

For block A

10-F = 0.5*vf² - 0.5*4*2²

= 2vf²-8

10-f = 2vf²-8

Collect like terms

10+8-f = 2vf²

18-f = 2vf²

F = 2vf² - 18 -----(2)

Equate equation 1 and 2 together

0.5vf²-2 =  2vf² - 18

take like terms here

0.5vf² + 2vf² = 18+2

2.5vf² = 20

divide through by 2.5

vf² = 8

The magnitude of both A and B are said to be equal

Kinetic energy of A

0.5*4*2.828²

= 15.99 Joules

The Kinetic energy of B

0.5*1*2.828²

= 3.99 joules

Read more on kinetic energy here: https://brainly.com/question/25959744

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