Question

mass= 250g, final velocity= 5m/s initially velocity 2m/s work done? ​

Answer 1

$\bigstar$ Given

Mass = 250 g

Final velocity = 5 m/s

Initial velocity = 2 m/s

$\bigstar$ To Find

Work done

$\bigstar$ Solution:

We know that,

$\star$ Work Energy Theorem

$\longrightarrow \sf W=\Delta K.E$

Here,

W = work done

Δ K.E = change in kinetic energy

We know that,

$\sf \longrightarrow \Delta K.E=\dfrac{1}{2}m(v^{2}-u^{2})$

Hence,

$\sf \longrightarrow W=\dfrac{1}{2}m(v^{2}-u^{2})$

According to the question,

We are asked to find the work done

So, we must find "W"

Given that,

Mass = 250 g

Final velocity = 5 m/s

Initial velocity = 2 m/s

Hence,

• m = 250 g = 0.25 kg
• v = 5 m/s
• u = 2 m/s

Substituting the values,

We get,

$\sf \longrightarrow W=\dfrac{1}{2} \times 0.25 \times (5^{2}-2^{2})$

$\sf \longrightarrow W=\dfrac{1}{2} \times 0.25 \times (25-4)$

$\sf \longrightarrow W=\dfrac{1}{2} \times 0.25 \times (21)$

On further simplification,

We get,

$\sf \longrightarrow W=2.625 \ J$

Answer 2

Given:-

• Mass = 250 g

• Final velocity = 5 m/s

• Initial velocity = 2 m/s

To Find:-

• Work done

$★ \sf Solution:-$

Formula :

Work Energy Theorem :-

↝ W=ΔK.E

Where,

W = work done

Δ K.E = change in kinetic energy

Also ,

$\purple{\sf ➝\Delta K.E=\dfrac{1}{2}m(v^{2}-u^{2})}$

Hence,

$\sf \longrightarrow W=\dfrac{1}{2}m(v^{2}-u^{2})$

Now,

m = 250 g = 0.25 kg

v = 5 m/s

u = 2 m/s

Substituting the values,

$\sf \implies W=\dfrac{1}{2} \times 0.25 \times (5^{2}-2^{2})$

$\sf \implies W=\dfrac{1}{2} \times 0.25 \times (25-4)$

$\sf \implies W=\dfrac{1}{2} \times 0.25 \times (21)$

Therefore,

$\sf \bigstar \boxed{\pink{W=2.625} }\ J$

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