HW4-phy2048-Spring-2014-solution

0 ms2 80 ms2 30 m2 47 m 6 3 phy2048

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Unformatted text preview: 0.935 m/s. 7P31. (a) As the body moves along the x axis from xi = 3.0 m to xf = 4.0 m the work done by the force is 1 2 xf xf xi 2 1 xi W = ∫ Fx dx = ∫ −6 x dx = −3( x 2 − xi2 ) = −3 (4.02 − 3.02 ) = −21 J. f According to the work- kinetic energy theorem, this gives the change in the kinetic energy: 1 W = ΔK = m v 2 − vi2 f 2 where vi is the initial velocity (at xi) and vf is the final velocity (at xf). The theorem yields 2W 2 2(−21 J) vf = + vi = + (8.0 m/s)2 = 6.6 m/s. m 2.0 kg (b) The velocity of the particle is vf = 5.0 m/s when it is at x = xf. The work- kinetic energy theorem is used to solve for xf. The net work done on the particle is W = −3 x2 − xi2 , so the theorem leads to f ( ( ) ( ) −3 x 2 − xi2 = f Thus, xf = − ) ( ) 1 m v 2 − vi2 . f 2 m2 2 2.0 kg (v f − vi ) + xi2 = − 6 N/m ((5.0 m/s)2 − (8.0 m/s)2 ) + (3.0 m)2 = 4.7 m. 6 3 PHY2048 Spring 2014 HW4 7P37. (a) We first multiply the vertical axis by the mass, so that it becomes a graph of the applied force. Now, adding the triangular and...
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