A disc of mass 100 ~g slides down from rest on an inclined plane of 30^ and come to rest after travelling a distance of 1 ~m along the horizontal plane. If the coefficient of friction is 0.2 for both inclined and horizontal planes, then the work done by the frictional force over the whole journey, approximately, is (Acceleration due to gravity, g=10 ~ms ⁻¹ )

Step-by-step solution

According to the question, acceleration of a block sliding down an inclined plane is shown in the following figure. From third equation of the motion, velocity of disc when it leaves the inclined plane, v^2=u^2-2 a s or u= 2 a s ( v=0) Acceleration of a block on a horizontal, gathered a= g . a=0.2 10=2 ~m / s ^2 (Given, =0.2 ) gathered After putting the value of a in Eq. (i), we get array rlrl & u & = 2 2 1 & =2 ~m / s array ( s=1 ~m , given ) From above diagram, the frictional force applied on the disc inclined pl