1.

A bolck of mass `10 kg` is moving in x-direction with a constant speed of `10 m//s`. it is subjected to a retardeng force `F=-0.1 x J//m`. During its travel from `x =20m` to `x =30 m`. Its final kinetic energy will be .A. 250 JB. 275 JC. 450 JD. 475 J

Answer» Correct Answer - D
Here, m=10 kg, `v_i= 10 m s^(-1)`
Initial kinetic energy of the block is
`K_i=1/2mv_i^2=1/2xx(10 kg)xx(10 m s^(-1))^2 =500 J`
Work done by retarding force
`W=intF_rdx=underset(20)overset(30)int -0.1 xdx = -0.1 [x^2/2]_20^30`
`= -0.1[(900-400)/2]= -25 J`
According to work energy theorem
`W=K_f-K_i`
`K_f=W+K_i = - 25 J + 500 J = 475 J `


Discussion

No Comment Found

Related InterviewSolutions