(a) If the particle can reach x = 1.0m, what is its speed there, and if it canno

Question

(a) If the particle can reach x = 1.0m, what is its speed there, and if it cannot, what is its turningpoint?
m or m/s

(b) What are the magnitude and direction of the force on theparticle as it begins to move to the left of x = 4.0m?
N in the Negative orPositive direction. (c) If the particle can reach x = 7.0m, what is its speed there, and if it cannot, what is its turningpoint?
m or m/s

(d) What are the magnitude and direction of the force on theparticle as it begins to move to the right of x = 5.0m?
N in the Negative orPositive direction.

Explanation / Answer

Total energy = KE + PE = E = 25 + 1/2 m v2 = 25 +1.32 * 64 / 2 = 67.2 J    Since forces are conservative the total energy is always 67.2J a) E1 = 67.2 - 40 = 27.2J       kinetic energy at x = 1 1/2 m v2 =27.2        v = (54.4 /1.32) = 6.42 m/s b) F = - V / x = 15 / 2 = 7.5N             (V and x have different signs)     F is postive and directed to theright c)   Again E = 67.2 J It's KE at x = 7 is 67.2 - 55 = 12.2 J      v = (2 * 12.2 / 1.32) = 4.30m/s d) F = - V / x = - 30 / 1 = - 30 N

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