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goblinko [34]
2 years ago
6

A certain spring has a spring constant k1 = 660 N/m as the spring is stretched from x = 0 to x1 = 35 cm. The spring constant the

n changes to k2 = 250 N/m as the spring is stretched to x2 = 65 cm. From x2 = 65 cm to x3 = 89 cm the spring force is constant at F3 = 105 N.
Write an equation for the work done in stretching the spring from x1 to x2.
Calculate the work done, in joules, in stretching the spring from x1 to x2.
Calculate the work, in joules, necessary to stretch the spring from x = 0 to x3.
Physics
1 answer:
pantera1 [17]2 years ago
6 0

(a) The equation for the work done in stretching the spring from x1 to x2 is ¹/₂K₂Δx².

(b) The work done, in stretching the spring from x1 to x2 is 11.25 J.

(c) The work, necessary to stretch the spring from x = 0 to x3 is 64.28 J.

<h3>Work done in the spring</h3>

The work done in stretching the spring is calculated as follows;

W = ¹/₂kx²

W(1 to 2) = ¹/₂K₂Δx²

W(1 to 2)  =  ¹/₂(250)(0.65 - 0.35)²

W(1 to 2)  = 11.25 J

W(0  to 3) = ¹/₂k₁x₁² + ¹/₂k₂x₂² + ¹/₂F₃x₃

W(0  to 3) = ¹/₂(660)(0.35)² + ¹/₂(250)(0.65 - 0.35)² + ¹/₂(105)(0.89 - 0.65)

W(0  to 3) = 64.28 J

Learn more about work done here: brainly.com/question/25573309

#SPJ1

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A 4 kg toy car moves horizontally on a rough road with coefficient of kinetic friction 0.2. It accelerates from rest to 20 m/s i
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The total work done on the car is 784Joule.

<h3>What's the acceleration of the car?</h3>
  • As per Newton's equation of motion, V= U+at
  • U= initial velocity= 0 m/s

V= vinal velocity= 20m/s

t= time = 10s

a= acceleration

  • So, 20= 0+ 10a

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<h3>What's the distance covered by the car in 10 seconds?</h3>
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V²-U² = 2aS

  • S= distance covered by the car
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<h3>What's the work done on the car due to frictional force?</h3>

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