An object of mass 'm' is being moved with a constant velocity under the action of an applied force of 2 N along a frictionless surface with following surface profile.
The correct applied force vs distance graph will be :
The correct applied force vs distance graph will be :
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The Correct Option is C
Solution and Explanation
On the rising incline (first half of the distance D): the applied force acts up the slope and balances \(mg\sin\theta\), so \(F = +2\) N (given).
On the descending incline (second half): gravity's component \(mg\sin\theta\) now acts along the motion, so the applied force must act opposite to the motion to keep the speed constant: \(F = -mg\sin\theta = -2\) N (the two slopes have the same angle \(\theta\), so the magnitude is the same).
Hence F is \(+2\) N for the first \(D/2\) and \(-2\) N for the next \(D/2\): two equal rectangular steps, positive then negative. This is graph (C), the third option.
Answer: (3) graph with +2 N for the first half and -2 N for the second half
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