An inclined plane making an angle (30^ ) with the horizontal is placed in a uniform horizontal electric field of (100 Vm ⁻¹ ) as shown in the figure. A small block of mass ( I kg ) and charge, 0.01C is allowed to slide down from rest from a height, (h=1 ~m ). If the coefficient of friction is 0.2 , then the acceleration of the block is nearly, (Acceleration due to gravity, (g=10 ~ms ⁻² ) )

Step-by-step solution

According to the question, an inclined plane is making an angle of (30^ ) with the horizontal, placed in a uniform electric field of (100 ~nm ⁻¹ ). It can be such in the figure that a block of mass (m ) is sliding down from rest at height (h ). From the above diagram, the total force (F ) acting along inclined plane. From fig, (m g 30^ - m g 30^ -q E 30^ =m a=F ) Given, ( =0.2, m=1 ~kg ,, q=0.01 C ) and (h=1 ~m ) Putting these values, we get ( aligned & 10 1 2 -0.2 10 3 2 -0.01 100 3 2 =a & a=5- 3 -0.5 3 2.3 ~ms ⁻²