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

Answer

D

Explanation

F=-0.1`xx`J/m
According to Work Energy theorem
Work done by all force = `K_t-K_i`
`implies int F.dx=K_t-k_i`
`implies overset(30)underset(20)(int)-0.1xxdx=K_t-1/2xx mu^2`
`(-)0.1[(x^2)/(2)]_(20)^(30)=K_t-1/2xx10xx10^2`
`(1)/(10xx2)[x^2]_(30)^(20)=K_t-500`
Limit inverse to make -ve to positive
`(1)/(20)xx[400-900]=K_t-500`
`-(500)/(20)=K_t-500`
`K_t=500-25=475` J