A block of mass 10 kg, moving in x-direction with a constant speed of `10ms^(-1)`, is subjected to a retarding force `F=0.1xxJ//m` during its travel from x=20 m to 30 m. Its final KE will be

Options

A. 475 J
B. 450 J
C. 275 J
D. 250 J

Answer

A

Explanation

Initial K.E. `E_1=1/2mv^(2)`
`=1/2xx10xx(10)^(2)=500J`
At x=20m, retarding force,
`F_(1)=0.1xx20=2N`
At x=30m, retarding force,
`F_(2)=0.1 xx 30=3N`
Average Retarding Force `F=(F_(1)+F_(2))/(2)=(2+3)/(2)=2.5`
Work done by retareding force =loss in K.E.
`=F xx x =2.5(30-20)=25J`
Final K.E. `E_(1)-"loss in K.E."=500-25=475j`