Total opposing force is 7 N plus 3 N, totaling 10 N, producing a deceleration of 5 m s-2 on the 2 kg mass. Using kinematics, the stopping distance traveled before coming to rest is 10 meters.
An object of mass 2 kg moving with a constant velocity of 10 m s-1 encounters a rough patch where the force of friction on the object is 7 N. At the same time, an additional constant force of 3 N opposing the motion is applied on the object. After entering the rough patch, how much distance does the object travel before coming to rest?
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Both the frictional force of 7 N and the additional force of 3 N oppose motion, producing a total retarding net force of 10 N. By Newton’s second law, dividing 10 N by 2 kg gives an acceleration of minus 5 m s-2. Using the kinematic equation v squared equals u squared plus 2 a s, 0 equals 100 minus 10 s, yielding a distance of 10 meters.
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