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A particle of mass 2.0kg moves under the influence of the force F(x)=(−5x2+7x)N. If its speed at x=−4.0m is v=20.0m/s, what is its speed at x=4.0m?

a) 10.0 m/s
b) 12.0 m/s
c) 14.0 m/s
d) 16.0 m/s

1 Answer

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Final answer:

The speed of the particle at x = 4.0 m is approximately 15.5 m/s.

Step-by-step explanation:

To find the speed of the particle at x = 4.0 m, we can first calculate the force at that position using the given force function F(x) = (-5x^2 + 7x) N. Plugging in x = 4.0 m into the equation, we get F(4.0) = -80 N.

Since force is equal to mass times acceleration, we can use the formula F = ma to find the acceleration of the particle. Rearranging the equation, we have a = F/m. Plugging in the values, we get a = -80 N / 2.0 kg = -40.0 m/s^2.

Finally, we can use the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement, to find the speed at x = 4.0 m. Since the particle's initial speed at x = -4.0 m is given as 20.0 m/s, we can plug in the values to get v^2 = 20.0^2 + 2(-40.0)(4 - (-4)) = 400 + 2(-40)(8) = 400 - 640 = -240 m^2/s^2. Since speed cannot be negative, we take the absolute value of the result, giving us a speed of 15.5 m/s.

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