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In Physics / High School | 2014-11-21

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

Asked by kurtisb76

Answer (3)

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

Answered by adamrplonski | 2024-06-10

Work done by the force of friction 900 J and coefficient of friction Us=0.166 ;

Answered by davidlcastiblanco | 2024-06-12

The work done by the force of friction on the cart is -900 J, indicating that the friction force opposes the motion. This was calculated using Newton's laws to find the net force and the force of friction before applying the work equation. The result demonstrates the role of friction in reducing the energy of the cart's motion.
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Answered by davidlcastiblanco | 2024-12-26

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