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3 years ago

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If the elevator cable breaks when the elevator is at a height h above the top of the spring, calculate the value that the spring

constant k should have so that passengers undergo an acceleration of no more than 7.0 g when brought to rest. Let M be the total mass of the elevator and passengers.

1 answer:

lawyer [7]3 years ago

4 0

Answer:

$k=\frac{24Mg}{h}$

Explanation:

We have to know the maximum compression of the spring in order to know the total height at which the elevator falls. According to Newton's second law:

$\sum F=ma\\kx-Mg=M(7g)\\kx=8Mg\\x=\frac{8Mg}{k}$

The law of the conservation of energy states that:

$MgH=\frac{kx^2}{2}\\Mg(h+x)=\frac{k(\frac{8Mg}{k})^2}{2}\\Mgh+Mg(\frac{8Mg}{k})=\frac{64M^2g^2}{2k}\\Mgh=\frac{32M^2g^2}{k}-\frac{8M^2g^2}{k}\\k(Mgh)=24M^2g^2\\k=\frac{24Mg}{h}$

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2 years ago

Two small nonconducting spheres have a total charge of 90.0 C.

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Answer: (a) Smaller charge is $2.7 \times 10^{-5} C$ and larger charge is $11.7 \times 10^{-5} C$ .

(b) Smaller charge is $-11.4 \times 10^{-5}$ and larger charge is $9.1 \times 10^{-5}$ .

Explanation:

(a) When both the spheres have same charge then force is repulsive in nature as like charges tend to repel each other.

Therefore, total charge on the two non-conducting spheres will be calculated as follows.

$Q_{1} + Q_{2} = 90 \mu \times \frac{10^{-6}C}{1 \muC}$

= $9 \times 10^{-5} C$

Therefore, force between the two spheres will be calculated as follows.

F = $k\frac{Q_{1}Q_{2}}{r^{2}}$

12 N = $\frac{(9 \times 10^{9} Nm^{2}/C^{2})Q_{1}Q_{2}}{(0.28 m^{2})}$

$Q_{1}Q_{2} = 0.104 \times 10^{-9} C^{2}$

or, $Q_{1}(9 \times 10^{-5} - Q_{1}) = 0.104 \times 10^{-9} C^{2}$

$9 \times 10^{-5}Q_{1} - Q^{2}_{1} = 0.104 \times 10^{-9} C^{2}$

$Q^{2}_{1} - 9 \times 10^{-5}Q_{1} + 0.104 \times 10^{-9} = 0$

$Q_{1} = 11.7 \times 10^{-5} C, 2.7 \times 10^{-5} C$

This means that smaller charge is $2.7 \times 10^{-5} C$ and larger charge is $11.7 \times 10^{-5} C$ .

(b) When force is attractive in nature then it means both the charges are of opposite sign.

Hence, total charge on the non-conducting sphere is as follows.

$Q_{1} + (-Q_{2}) = 90 \mu \times \frac{10^{-6}C}{1 \muC}$

$Q_{1} - Q_{2} = 9 \times 10^{-5} C$

Now, force between the two spheres is calculated as follows.

F = $k\frac{Q_{1}Q_{2}}{r^{2}}$

12 N = $\frac{(9 \times 10^{9} Nm^{2}/C^{2})Q_{1}Q_{2}}{(0.28 m^{2})}$

$Q_{1}Q_{2} = 0.104 \times 10^{-9} C^{2}$

$Q_{1}(Q_{1} - 9 \times 10^{-5}) = 0.104 \times 10^{-9} C^{2}$

$Q^{2}_{1} - 9 \times 10^{-5}Q_{1} = 0.104 \times 10^{-9} C^{2}$

$Q_{1} = -11.4 \times 10^{-5}, 9.1 \times 10^{-5}$

Hence, smaller charge is $-11.4 \times 10^{-5}$ and larger charge is $9.1 \times 10^{-5}$ .

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