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A 26 g particle is moving to the left at 13 m per s How much net work must be done on the particle to cause it to move to the right at 49 m per s?
Wiki User
∙ 14y ago
Final kinetic energy will be equal to (Initial kinetic energy) plus (work that you do to turn it around).
KE = 1/2 m V2
Initial KE = 1/2 (0.026) (13)2 with 'V' pointing left.
Final KE = 1/2 (0.026) (49)2 with 'V' pointing right.
You have to supply the difference = 1/2 (0.026) (62)2 = 49.972 kg-m2/s2 = 49.972 newton-m
= 49.972 joules
750 J
537.5 J
Wiki User
∙ 14y ago
The change in velocity is (\Delta v = 49 , m/s - (-13 , m/s) = 62 , m/s). The work done is given by (W = \frac{1}{2} m (\Delta v)^2), where (m) is the mass of the particle. Substituting the given values, the net work required is (W = \frac{1}{2} \times 26 , g \times (62 , m/s)^2).
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