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You apply a constant force F=(68.0N)i^+(36.0N)j^ to a 420kg car as the car travels 58.0m in a direction that is 240.0 counterclockwise from the +x-axis. How much work does the force you apply do on the car?
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Added Tue, 04 Aug '15
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Work done is equal to dot product of Force and displacement vectors. . Here, the force is F=(68.0N)i^+(36.0N)j^ and you need to write the displacement vector . As the displacement makes an angle of 60 degrees from negative x axis and 30 degrees from negative y axis. So the displacement vector would be (-58cos60i^-58cos30j^)m. Now, the dot poduct of these two vectors would be ((68)*(-58cos60)+(36)*(-58cos30))N-m. This when done equals to -3778.12 Joules.
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Added Tue, 04 Aug '15
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