A body of mass 5g is moving in a straight line with a velocity of `10 cm *s^(-1)`. A force of `10 sqrt(2)` dyn is applied on the body at an angle of `45^@` with the line of motion. What is the change in kinetic energy of the body in 1 st second ?

Explanation

Let the line of motion of the body be chosen as the x-axis.
Initial kinetic energy of the body
`=1/2 mv^2 =1/2 xx5xx(10)^2=250 erg`
Components of the applied force along and at right angle to the direction of motion are respectively,
`F_x=10sqrt(2) cos 45 ^2 =10 sqrt(2)*1/sqrt(2) =10 ` dyn
`and F_y =10 sqrt(2) sin 45^@ =10 sqrt(2) *1/sqrt(2) =10` dyn
`therefore` Components of acceleration of the body,
`a_x=a_y=(F_x)/(m) =(F_y)/(m) = (10)/(5) =2 cm *s^(-2)`
`therefore` The velocity along x-axis after 1 s `v_x=10+2xx1= 12 cm *s^(-1)` and along y-axis after 1 s, `v_y =0+2xx1= 2 cm *s^(-1)`
Total kinetic energy of the body after 1 s
`=1/2 m(v_x^2+v_y^2)`
`=1/2 xx5xx(12^2+2^2) =1/2 xx5xx148 =370 erg`
`therefore` Change in kinetic energy in 1 second = 370 -250 =120 erg.