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brilliants [131]
2 years ago
10

As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff springs arr

anged side by side so that they are parallel to each other. When you push the platform, you compress the springs. You do an amount of work of 76.0 J when you compress the springs a distance of 0.170 m from their uncompressed length.
(a) What magnitude of force must you apply to hold the platform in this position?
(b) How much additional work must you do to move the platform 0.200 m farther, and what maximum force must you apply?
Physics
1 answer:
zubka84 [21]2 years ago
8 0

Answer:

894.12\ \text{N}

284.013\ \text{J}

Explanation:

U = Energy = 76 J

x = Displacement of spring = 0.17 m

k = Spring constant

Energy is given by

U=\dfrac{1}{2}kx^2\\\Rightarrow k=\dfrac{2U}{x^2}\\\Rightarrow k=\dfrac{2\times 76}{0.17^2}\\\Rightarrow k=5259.51\ \text{Nm}

Force is given by

F=kx\\\Rightarrow F=5259.51\times 0.17\\\Rightarrow F=894.12\ \text{N}

The magnitude of force must you apply to hold the platform is 894.12\ \text{N}.

Now x=0.17+0.2=0.37\ \text{m}

U=\dfrac{1}{2}kx^2\\\Rightarrow U=\dfrac{1}{2}\times 5259.51\times 0.37^2\\\Rightarrow U=360.013\ \text{J}

Additional energy

360.013-76=284.013\ \text{J}

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3 years ago
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3 years ago
A 6.75 nC charge is located 1.99 m from a 4.46 nC point charge.
sladkih [1.3K]

Explanation:

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Charge 1, q_1=6.75\ nC=6.75 \times 10^{-9}\ C

Charge 2, q_2=4.46\ nC=4.46\times 10^{-9}\ C

The distance between charges, r = 1.99 m

To find,

The electrostatic force and its nature

Solution,

(a) The electric force between two charges is given by :

F=\dfrac{kq_1q_2}{r^2}

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5 0
3 years ago
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