An inclined plane making an angle of 30° with the horizontal is placed in a uniform horizontal electric field 200N/C as shown in the figure. A body of mass 1 kg and charge 5 mC is allowed to slide down from rest at a height of 1 m. If the coefficient of friction is 0.2, find the time taken by the body to reach the bottom.
[g = 9.8 m/s²; sin 30° = 1/2; cos 30° = √3/2]

Solution

Physics sir · 4 steps

The field pushes the block back up the slope and into it, so it slows the block and increases friction.

1. Electric force. qE = 5 × 10⁻³ × 200 = 1 N, horizontal and to the right. The block slides down to the left. Along the slope this force gives 1 × cos 30° = 0.866 N up the incline, and 1 × sin 30° = 0.5 N into the incline.
2. Normal and friction. Gravity presses with mg cos 30° = 9.8 × 0.866 = 8.487 N, and the field adds 0.5 N, so N = 8.987 N. Friction acts up the slope: f = 0.2 × 8.987 = 1.797 N.

   ma = mg sin 30° - qE cos 30° - μ(mg cos 30° + qE sin 30°)

3. Acceleration. a = 4.9 - 0.866 - 1.797 = 2.237 m s⁻² down the slope. The slope length is h/(sin 30°) = 1/0.5 = 2 m.
4. So. Starting from rest, s = 1/2at² gives t = √((2 × 2)/2.237) = √1.788 ≈ 1.34 s, which matches 1.3 s, option (c).

The idea to keep. Resolve every extra force along the slope and normal to it. The normal part changes N, and so it changes friction too. Then use the slope length, h/sin θ, in the kinematics.

Answer from the NTA final answer key. Solution by PrepWiser.