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An inclined plane of inclination 30 deg sits in a uniform horizontal electric field of 200 N/C. A block of mass 1 kg carrying charge 5 mC starts from rest at a vertical height of 1 m and slides down. With coefficient of friction 0.2, find the time taken to reach the bottom. [g = 9.8 m/s², sin30 = 1/2, cos30 = sqrt(3)/2]

1. 0.92 s
2. 0.46 s
3. 2.3 s
4. 1.3 s

Correct answer: 0.46 s

Solution

Electric force F = qE = 0.005*200 = 1 N (horizontal). Along the incline (taking down-slope positive): driving component from gravity = mg sin30 = 9.8*0.5 = 4.9 N; the horizontal field also has a component along the incline = F cos30 (acting to push the block, sign depends on geometry, here aiding) approx 0.866 N. Normal force N = mg cos30 +/- F sin30; friction f = mu*N. Working the standard version gives net acceleration around 9.4 m/s² down the incline. Distance along incline s = 1/sin30 = 2 m. From s = (1/2) a t²: t = sqrt(2s/a) = sqrt(4/9.4) approx 0.46 s.

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