(d) What are the magnitude and direction of the force on the particle as it begi
ID: 1962477 • Letter: #
Question
(d) What are the magnitude and direction of the force on the particle as it begins to move to the right of x = 5.0 m?
______ N in the negative direction.
hint: The mechanical energy (the sum of K and U) remains constant as the particle moves. The graph gives U. You can thus calculate K, which cannot be negative. From K, you can calculate speed. A turning point is reached if K goes to zero. There a particle stops and reverses its motion. The force slowing, stopping, and reversing the particle is equal to the negative of the graph's slope.
Explanation / Answer
KE = 1/2 * .86 * 7.5^2 = 24.2 J KE of particle when released U3 - U1 = 55 - 20 = 35 J Potential energy gained moving from U1 to U3 The particle does not have enough KE to reach x = 7 Slope from x = 5 to x = 6 is (U3 - U1) / 1 = 35 J / m X = 24.2 / 35 since (using d for delta) dU / dx = 35 and dU = 35 * dx Since dU = 24.2 for all of the KE to be expended dx = 24.2 / 35 = .67 Then the particle will stop at X = 5 + .67 = 5.67 m F = -dU/dx = -35 N (in the negative direction) Note J = N-m and J/m = N for the units involved
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