An inclined plane of length 5.60 m, making an angle of 45° with the horizontal, is placed in a uniform, vertically upward electric field E = 100 V m⁻¹. A particle of mass 1 kg and charge 10⁻² C is allowed to slide down from rest from the top of the slope. If the coefficient of friction is 0.1, the…

Question

An inclined plane of length 5.60 m, making an angle of 45° with the horizontal, is placed in a uniform, vertically upward electric field E = 100 V m⁻¹. A particle of mass 1 kg and charge 10⁻² C is allowed to slide down from rest from the top of the slope. If the coefficient of friction is 0.1, the time taken by the particle to reach the bottom is . (Take g = 9.8 m s⁻².)

Accepted Answer

(2) 1.41 s. Given: L = 5.60 m, θ = 45°, μ = 0.1, m = 1 kg, q = 10⁻² C, E = 100 V m⁻¹ upward, g = 9.8 m s⁻². Idea: the upward electric force qE just reduces the particle's effective weight; after that it is an ordinary rough incline. qE = (10⁻²)(100) = 1 N, so the effective weight is mg - qE = 9.8 - 1 = 8.8 N, an effective g' = 8.8 m s⁻². Normal reaction N = mg'cos 45° and friction μ N, so along the incline a = g'(sin 45° - μ cos 45°). a = 8.8×(1 - 0.1)/(√2) = (7.92)/(1.414) = 5.6 m s⁻² From rest, L = 1/2at², so t = √(2L)/a = √(2×5.6)/(5.6) = √2 s. So t = 1.41 s.

Step-by-step solution

Why the other options are wrong