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2 years ago

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A block weighing 15 newtons is pulled to the top of an incline that is 0.20 meter above the ground, as shown below. if 4.0 joule

s of work are needed to pull the block the full length of the incline, how much work is done against friction?

1 answer:

Annette [7]2 years ago

3 0

Total work done here against two forces

1. Work done against gravity

2. work done against friction

Here work done against gravity is given by

$W_g = mgH$

here we know that weight of the block will be

$Weight = mg = 15 N$

$H = 0.20 m$

$W_g = 15(0.20) = 3J$

now we will say

$W_{total} = W_{friction} + W_g$

as we know that total work

$W_{total} = 4J$

$4 = 3 + W_{friction}$

$W_{friction} = 4 - 3 = 1J$

so work done against friction is 1 J

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Answer:

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m is given and we find k from the equilibrium extension of 6.0 cm (0.06 m):

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k*y - W = 0 ⇒ k*y - m*g = 0 ⇒ k = m*g / y

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⇒ b = 2*√(k*m) = 2*√(408.75 N/m*2.5 kg)

⇒ b = 63.9335 Kg/s ≈ 64 Kg/s

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Use Wien’s Law to calculate the peak wavelength of Betelgeuse, based on the temperature found in Question #8. Note: 1 nanometer

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The peak wavelength of Betelgeuse is 828 nm

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Answer:

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Answer:

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