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n200080 [17]
3 years ago
13

a boy is riding a bicycle at a velocity of 5.0 m/s. he applies the brakes and uniformly decelerates to a stop at a rate of 2.5 m

/s. how long does it take for the bicycle to stop.
Physics
1 answer:
JulijaS [17]3 years ago
3 0

The working equation would be Vf (final velocity) = Vi (initial velocity) + a (acceleration) t (time). The given data are the initial velocity (5.0 m/s), acceleration (-2.5 m/s^2, negative since it is said to decelerate) and the final velocity (0 m/s, since it will put to a stop). The time would be 2 seconds. 

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The work done by the charge on the origin to the moving charge is equal to the difference in the potential energy of the charges.

This is the electrostatic equivalent of the work-energy theorem.

W = \Delta U = U_2 - U_1

where the potential energy is defined as follows

U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}

Let's first calculate the distance 'r' for both positions.

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Now, we can calculate the potential energies for both positions.

U_1 = \frac{kq_1q_2}{r_1^2} = \frac{(8.99\times 10^9)(2.3\times 10^{-6})(-5\times 10^{-6})}{(0.170)^2} = -3.57~J\\U_2 = \frac{kq_1q_2}{r_2^2} = \frac{(8.99\times 10^9)(2.3\times 10^{-6})(-5\times 10^{-6})}{(0.3530)^2} = -0.829~J

Finally, the total work done on the moving particle can be calculated.

W = U_2 - U_1 = (-0.829) - (-3.57) = 2.74~J

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