A force of 4N acts on a body of mass 2kg for 4s. Assuming the body to be initially at rest, find

a) its velocity when the force stops acting.

b) the distance covered in 10s after the force starts acting.

F = m(dv/dt)

so,

dv = (F/m).dt

here

F = 4N

m = 2kg

dt = 4s

thus, the velocity will be

dv = (4/2).4

so,

dv = 8 m/s

acceleration

a = dv/dt = 8/4 = 2 m/s^{2}

distance covered while force is acting

s_{1 }= (v^{2} - u^{2}) / 2a

= (8^{2} - 0) / 2x2

or

s_{1} = 16 m

..

(b)

distance covered while force is not acting

s_{2} = ut + (1/2)at^{2}

here,

u = 8 m/s

t = 10s - 4s = 6s

a = 0

so,

s_{2} = 8x6 = 48m

thus,

the total distance travelled

S = s_{1} + s_{2} = 16m + 48m

or

**S = 64m**

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