Question

A body of mass 5kg is moved along a straight line and its displacement c varies with time t according to relation t=√x+1 find work done in moving the body for 1st 10 sec

Answer 1

Answer:

Mass of the body = 5 kg

A/c to the question, t=√x+1

For 10 sec, 10=√x +1

I.e. X=81

For 0 sec, 0=√x+1

I.e. X=1

So, work done ,W= FS

Or, mas

Or, 5 kg× 8ms^-2 ×10s

Or, 400, the answer

Explanation:

Answer 2

Answer:

800 J

Explanation:

Here, t= √x + 1

> √x = t-1

> x= t^2-2t +1

differentiating both sides w. r. t. t, we have

dx/dt =2t-2

> v= (2t-2) m/s

again differentiating both sides, we have

dv/dt = 2

> a= 2 m/s^2

therefore force applied to the body,

F= Ma= 5 * 2= 10 N

now, initial displacement = 0^2-2*0+1=1m

displacement of the body during 10 seconds

x= 10^2-2*10+1=81 m

therefore, displacement of the body in first 10 seconds

S= 81-1= 80 m

hence work done = 10*80= 800 J