A block of mass 5 kg is moving with a speed of 8 m/s. A force of 10 N is applied on it for 3 s . Find the final speed of the block if the force is applied (i) along the motion , (ii) opposite to the motion and (iii) normal to the intial direction of motion of the block.

Explanation

The acceleration produced by the force has magnitude , ` a = F/m = 2 m//s^(2)`
Time for which acceleration exists is 3 s.
(i) When force acts in the direction of motion ,
` u= 8 m//s , a = 2 m//s^(2) , t = 3 s`
` v= u+ at`
` v= 8+ 2 xx 3 = 14 m//s`
(ii) When force acts opposite to the direction of motion,
` u = 8 m//s , a = -2 m//s^(2) , t = 3 s`
` As v = u+ at `
` :. v = 8 - 2 xx 3 = 2 m//s`
(iii) When force acts perpendicular to the direction of motion, let the block be moving along x - axis and acceleration be aong y- axis . So, we have
` vec(u) = (8 m//s)hat(i) , a = ( 2 m//s^(2)) hat(j) , t = 3 s `
As `vec(v) = vec(u) + vec(a)t`
` :. vec(v) ( 8 m//s) hat (i) + (6 m//s)hat(j)`
` rArr "Speed" = |vec(v)| = sqrt(8^(2) + 6^(2))= 10 m//s `