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Natalka [10]
3 years ago
15

A massless spring with force constant ????=200N/m hangs from the ceiling. A 2.0-kg block is attached to the free end of the spri

ng and released. If the block falls 17 cm before starting back upwards, how much work is done by friction during its descent?
Physics
1 answer:
Makovka662 [10]3 years ago
8 0

Answer:

-0.4454 Joules

Explanation:

m = Mass of block = 2 kg

h = Height of extension = 17 cm = x

g = Acceleration due to gravity = 9.81 m/s²

Potential energy of the spring

P=mgh\\\Rightarrow P=2\times 9.81\times 0.17\\\Rightarrow P=3.3354\ J

The kinetic energy of the spring

K=\frac{1}{2}mx^2\\\Rightarrow K=\frac{1}{2}\times 200\times 0.17^2\\\Rightarrow K=2.89\ J

In this system as the potential and kinetic energy is conserved from work energy equivalence we get

W=P-K\\\Rightarrow W=2.89-3.3354\\\Rightarrow W=-0.4454\ J

The work done by friction is -0.4454 Joules

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Well, first of all, one who is sufficiently educated to deal with solving
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If the mass of the man standing on the weighing machine is 60kg, then
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The vertical speed of the deck is     y'(t) = M (2π/15) cos(2π t/15)

and its vertical acceleration is          y''(t) = - (2πM/15) (2π/15) sin(2π t/15)

                                                                = - (4 π² M / 15²)  sin(2π t/15)

                                                                = - 0.1755 M sin(2π t/15) .

There's the important number ... the  0.1755 M.
That's the peak acceleration.
From here, the problem is a piece-o-cake.

The net vertical force on the intrepid sailor ... the guy standing on the
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At the instant of peak acceleration, the weighing machine thinks that
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Now I'm going to wave my hands in the air a bit:

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The 'M' is what we need to find.

Divide each side by  0.1755 :          M = (0.08333 / 0.1755) G

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