Question

an object of mass 4kg accelerates from 3m/s to 6m/s in 8s.how much work is done on it?​

Answer 1

→ Question : An object of mass 4kg accelerates from 3m/s to 6m/s in 8s.  How much work is done on it?​

→ Answer: Work ( W ) = 54 Joules

→ Formula used :

    $F=ma\\\\a=\frac{\Delta V}{t}=\frac{v-u}{t} \\\\v^2-u^2=2as\\\\W=F.s$

→ Given :

mass , m = 4 kg

Initial velocity , u = 3 m/s

Final Velocity , v = 6 m/s

Time , t = 8 seconds

Work , W = ?

Explanation:

→ Here first we need to use Newton's second law, to find force.

=> F = ma

The acceleration a is the rate of change of velocity.

So,

   $a=\frac{\Delta V}{t}=\frac{v-u}{t}\\\\a=\frac{6-3}{8}=\frac{3}{8}=0.375 m/s^2$

=> F = 4 * ( 0.375 )

=> F = 1.5 N

The work done is force,F in the direction of s.

To find s , we need to use formula

     $v^2-u^2=2as\\\\=>v^2=u^2+2as\\\\=>(6)^2=(3)^2+2(0.375)(s)\\\\=>36=9+0.75s\\\\=>0.75s=27\\\\=>s=36 m$

So , work (W) = F * s

=> W = 1.5 * 36

=> W = 54 Joules

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Answer 2

$\bigstar$Answer:

$\star$Given:

Mass (m) = 4 kg

Accelerates from 3 m/s to 6 m/s

Therefore, ΔV = 6 -3 = 3 m/s

Time (t) = 8 s

$\star$To Find:

Work done (W)

$\star$Solution:

Using Newton's Second law

$\boxed{\rm F = ma}$

Given that,

Accelerates from 3 m/s to 6 m/s

Since, Acceleration is defined as the rate of change of velocity at given time

Therefore,

$\rm a=\Delta V = \dfrac{6-3}{8}\;m/s^2$

$\rm \implies a = 0.375\;m/s^2$

Hence,

$\boxed{\rm a =0.375\;m/s^2}$

Substituting it in F = ma

We get

$\implies \rm F = 4 \times 0.375$

$\rm \implies F = 1.5\;N$

$\boxed{\rm F = 1.5\;N}$

We know that,

Force x distance travelled along force gives Work done

$\boxed{\rm W = F \times s}$

So, We should find Distance (s) in order to find out Work done

Using Laws of motion,

$\boxed{\rm v^2 = u^2+2as}$

Given that,

v = 6 m/s

u = 3 m/s

And we found a = 0.375 m/s²

Substituting the above values in the given formula

We get,

$\implies \rm 6^2=3^2+(2 \times 0.375 \times s)$

$\rm \implies 36=9+0.75 \times s$

$\rm \implies s=\dfrac{27}{0.75}\;m$

$\rm \implies s=36\;m$

$\boxed{\rm s =36\;m}$

According to the question,

We are asked to find Work done (W)

$\boxed{\rm W = F \times s}$

Given that,

F = 1.5 N

and we found s = 36 m

Therefore,

$\implies \rm W = 1.5 \times 36\;J$

$\rm \implies W = 54\;J$

$\boxed{\rm W = 54\;J}$

Hence,

Work done is 54 J