Saved Bookmarks
| 1. |
A body of mass 0.3 kg is taken up an inclined plane of length 10 m and height 5 m, and then allowed to slide down the bottom again. The coefficient of friction between the body and the plane is 0.15. What is the (i) work done by the applied force over the upward journey? (ii) work the by gravitational force over the round trip? (iii)work the by the frictional force over the round trip? Which of the above forces (except applied force )is/are conservative forces? |
Answer» Solution :Upward journey - Let us calculate work done by different forces over upward journey. Work done by gravitational force, `W_(1)=(mg SIN theta)"s " cos 180^(@)` `W_(1)=0.3xx10" sin " 30^(@)xx10(-1)` `W_(1)=-15 J` Work done by force of FRICTION, `W_(2)=(mu " mg cos"theta)s cos 180^(@)` `W_(2)=0.15xx0.3xx10" cos"30^(@)xx10[-1]` `W_(2)=-3.879 J` Work done by external force, `W_(3)=F_(ext)xx s xxcos0^(@)` `W_(3)=[mg" sin"theta + mu "mg cos"theta]xx10xx1` `W_(3)`=18.897 J Downward journey Work done by the gravitational force, `W_(4)=mg" sin"30^(@)xx"s " cos0^(@)` `W_(4)=0.3xx10xx(1)/(2)xx10=-15 J` Work done by the FRICTIONAL force , `W_(5)=mu" mg " cos30xx"s cos" 180^(@)` `=0.5xx0.3(10sqrt3)/(2)xx10xx(-1)` =-3.897 (i) Work done by appllied force over upward journey, `W_(3)`=18.897 J (ii) Work done by gravitational force over the round trip `=W_(1)+W_(4)=0 J` (iii) Work done by the frictional force over the round trip, `W_(2)+W_(5)`=-3.897+(-3.897)=-7.794 J Work done by gravitational force over a closed path is zero but DUE to frictional force, it is non zero.Therefore,gravitational force is conservative and frictional force is non-conservative. |
|