Net work 2

A 2.0 kg 2.0 \text{ kg} particle with initial velocity 5 x ^ 4 y ^ m/s 5 \hat{x} - 4\hat{y} \text{ m/s} now has a velocity of 7 x ^ + 3 y ^ m/s. 7\hat{x} + 3\hat{y} \text{ m/s.} Assuming that no energy was lost in the process, how much work was done by the resultant force during the time interval?

x ^ and y ^ \hat{ x } \text{ and } \hat{ y} are unit vectors along x and y directions in a 2 2 -dimensional plane.

9 J 9 \text{ J} 34 J 34 \text{ J} 19 J 19 \text{ J} 17 J 17 \text{ J}

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2 solutions

Karan Joisher
Jul 5, 2014

u = √(5²+4²) = √41.... v= √(7²+3²) = √58.... v²= u² + 2as.... 58=41+2as.... a = 17/2s... F=ma=2a.... W=Fs=2as=17*2s/2s=17 ...

Amogh Jain
May 18, 2014

change in kinetic energy = work done

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