A body of mass 5 kg has momentum of 10 kg `ms^(-1)` when a force of 0.2 N is applied on it for 10 seconds , what is the change in the kinetic energy ?

Options

A. `1.1` J
B. `2.2`J
C. `3.3` J
D. `4.4` J

Answer

D

Explanation

`v_(0) =p/m = 10/5 = 2 ms^(-1)`
Acceleration `a = F/m = (0.2)/5 = 0.04 ms^(-2)`
` v = v_(0)+at`
` = 2+ 0.04 xx10`
` = 2+0.4`
` = 2.4 ms^(-1)`
` :. ` change in kinetic energy
`DeltaK = 1/2 mv^(2) - 1/2 mv_(0)^(2)`
` = 1/2 m(v^(2) -v_(0)^(2))`
`= 1/2 xx5(2.4^(2)-2^(2))`
` = 1/2 xx(5.76 -4.00)`
`= 1/2 xx5xx1.76 `
= 4.4 J