Saved Bookmarks
| 1. |
A block of mass `10 kg` is moving in x-direction with a constant speed of `10 m//s`. it is subjected to a retarding force `F=-0.1 x J//m`. During its travel from `x =20m` to `x =30 m`. Its final kinetic energy will be.A. `450J`B. `275J`C. `250J`D. `475 J` |
|
Answer» Correct Answer - D Here, `m=10kg, u=10 m//s, F=-0.1x J//m` `[`as force is retarding `]` `x_(1)=20cm, x_(2)=30cm` As `W=int_(x_(1))^(x_(2))Fdx=int_(20)^(30)-0.1x dx =-0.1[(x^(2))/(2)]_(20)^(30)` `=-(0.1)/(2)[(30)^(2)-(20)^(2)]=-(0.1)/(2)[900-400]` `=-(0.1)/(2)[500]=-25J` `K.E_(i)=(1)/(2)m u^(2)=(1)/(2)xx10xx(10)^(2)=500J` According to work-energy theorem, `W=K.E_(f)-K.E_(i)` or `-25=K.E_(f)-500` `K.E_(f)=475J` |
|