A force F = 3.51 N I + 5.30 N j acts on a panicle producing a displacement Delta

Question

A force F = 3.51 N I + 5.30 N j acts on a panicle producing a displacement Delta r = 8.36m Find the work done by the force on the particle. An archer pulls her bowstring buck 0.23 m, if the spring constant of the bow string is 510N/m, how much work does the archer do in pulling the bow? If you do not have a written valid excuse for missing the quiz, please also solve the following problem The kinetic energy of a 2.3 kg particle is 120j al point A. The speed of this panicle is 31 m/s when the particles reaches point B. What is the net work done on the panicle as it moves from A to B? W = F middot Delta r A middot B = AB cos theta A middot B = A_xB_x + A_y B_y + A_zB_z W_ext = DeltaK = 1/2 mvf^2 - 1/2 mv_i^2 W_exit = Delta U F_g = mg U_g = mg Delta h F_e = -kx U_x = 1/2 kx^2

Explanation / Answer

1) The work done by the force on the particle is

W = F * delta r

= (3.51 * 8.36) + (5.30 * 0)

= 29.34 J

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2) The work done by the archer through force F is equivalent to the elastic potential energy PE = 1/2 k d2

W = 1/2 * k d2

= 1/2 * 510 * 0.232

= 13.49 J

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