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Bess [88]
3 years ago
7

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 mov

ing 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.
Physics
1 answer:
Kitty [74]3 years ago
5 0
  <span>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</span>
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That is,

1/2(mv^2) = (K* q1q2)/r

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MakcuM [25]
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Find the distance traveled by a particle with position (x, y) as t varies in the given time interval. x = 2 sin2(t), y = 2 cos2(
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The distance traveled by the particle at the given time interval is 0.28 m.

<h3>Position of the particle at time, t = 0</h3>

The position of the particle at the given time is calculated as follows;

x = 2 sin2(t)

y = 2 cos2(t)

x(0) = 2 sin2(0) = 0

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<h3>Distance traveled by the particle at the given time interval</h3>

d = √[(x₄ - x₀)² + (y₄ - y₀)²]

d =  √[(0.28 - 0)² + (1.98 - 2)²]

d = 0.28 m

Thus, the distance traveled by the particle at the given time interval is 0.28 m.

Learn more about distance here: brainly.com/question/23848540

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