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andrey2020 [161]

3 years ago

9

How does work done on an object affect its kinetic energy

1 answer:

Alexus [3.1K]3 years ago

5 0

According to the work-energy theorem, the work done on an object by a net force equals the change in kinetic energy of the object. Essentially kinetic energy is the energy used for motion. Interestingly, as work is done on an object, potential energy can be stored in that object. A moving object has kinetic energy because work has been done on it. When work is done energy in one form is transferred to the kinetic energy of the moving object. To stop the object again, the same amount of work would have to be done to bring it back to rest.

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A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum

s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle= $q=-2.74\times 10^{-6} C$

Kinetic energy of particle= $K_E=6.65\times 10^{-10} J$

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We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

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Using the formula

$2.74\times 10^{-6}V_1=6.65\times 10^{-10} J$

$V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V$

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$\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2$

Using the formula

$\frac{V_1}{V_2}=(\frac{6.36}{t})^2$

$\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}$

$t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}$

$t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}$

$t=241.7 s$

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

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