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jasenka [17]
3 years ago
5

Work of 2 joules is done in stretching a spring from its natural length to 11 cm beyond its natural length. what is the force (i

n newtons) that holds the spring stretched at the same distance (11 cm)
Physics
1 answer:
Sidana [21]3 years ago
5 0
The law applied here is Hooke's Law which describes the force exerted by the spring with a given distance. The equation for this is F = kΔx, where F is the force in Newtons, k is the spring constant in N/m while Δx is the displacement in meters.

If you want to find work done by a spring, this can be solved by using differential equations. However, derived equations are already ready for use. The equation is

W = k[{x₂-x₁)² - (x₁-xn)²],

where 
xn is the natural length
x₁ is the stretched length 
x₂ is also the stretched length when stretched even further than x₁

In this case xn =x₁. So, that means that (x₁-xn) = 0 and (x₂-x₁) = 11 cm or 0.11 m.

Then, substituting the values,

2 J = k (0.11² -0²)
k = 165.29 N/m

Finally, we use the value of k to the Hooke's Law to determine the Force.

F = kΔx = (165.29 N/m)(0.11 m)
F = 18.18 Newtons
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Case d) has the strongest gravitational force

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The magnitude of the gravitational force between two objects is given by the equation:

F=G\frac{m_1 m_2}{r^2}

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G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2} is the gravitational constant

m1, m2 are the masses of the two objects

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r = 30 km = 30,000 m

So the gravitational force is

F=(6.67\cdot 10^{-11})\frac{(10)(2)}{30000^2}=1.48\cdot 10^{-18}N

b) For this pair of objects:

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m2 = 10 kg

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So the gravitational force is

F=(6.67\cdot 10^{-11})\frac{(10)(10)}{30000^2}=7.41\cdot 10^{-18}N

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F=(6.67\cdot 10^{-11})\frac{(2)(2)}{10000^2}=1.33\cdot 10^{-17}N

d) For this pair of objects:

m1 = 10 kg

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So the gravitational force is

F=(6.67\cdot 10^{-11})\frac{(10)(10)}{10000^2}=6.67\cdot 10^{-17}N

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