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18 September, 21:49

A horizontal force of 200N is applied to a 55kg cart across a 10m level surface. If the cart accelerates at 2.0m/s^2, then what is the work done by the force of friction as it slows the motion of the cart?

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  1. 19 September, 01:08
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    1j=1newton*meter

    force=mass*accel

    200N=55x

    200/55=3.636 ...

    3.636 ... - 2=1.636 ...

    1.636 is the deceleration resulting from friction hence the force of friction is 1.636*55=90newtons

    90newtons*distance of 10 meters = 900 j of work done by friction
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