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notsponge [240]
3 years ago
6

The spring is now compressed so that the unconstrained end moves from x=0 to x=L. Using the work integral W=∫xfxiF⃗ (x⃗ )⋅dx⃗ ,

find the work done by the spring as it is compressed. Express the work done by the spring in terms of k and L?

Physics
1 answer:
Savatey [412]3 years ago
4 0

solution:

the spring force exerted by spring with spring constant k is given by

F(x)=-kx

where k is spring constant

and x is deformation of spring

in order to calculate word done by the spring

W=\int\limits^L_0 {} \, dW

the work done by the spring as it is compressed from x=0 to x=L

W=-kx^2/2

inserting the limits x=0 and x=L

we get work done in terms of k and L

ANSWER

W=-kL^2/2

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pashok25 [27]
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By re-arranging this equation, we can find V_i:
V_i =  \frac{nRT_i}{p_i} = \frac{(2.3 mol)(8.31 J/mol K)(513 K)}{1.01 \cdot 10^5 Pa}=0.097 m^3

2) Now the gas cools down to a temperature of
T_f = 14^{\circ}C=287 K
while the pressure is kept constant: p_f = p_i = 1.01 \cdot 10^5 Pa, so we can use again the ideal gas law to find the new volume of the gas
V_f =  \frac{nRT_f}{p_f}= \frac{(2.3 mol)(8.31 J/molK)(287 K)}{1.01 \cdot 10^5 Pa} = 0.054 m^3

3) In a process at constant pressure, the work done by the gas is equal to the product between the pressure and the difference of volume:
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by using the data we found at point 1) and 2), we find
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A spring is used to stop a 50-kg package which is moving down a 20º incline. The spring has a constant k = 30 kN/m and is held b
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Rounding to 1 sig-fig, the spring is compressed an additional 0.3 meters.

8 0
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