Question

a body of mass 5 kg is initially at rest . suddenly it starts moving with acceleration of 10 m/s2 find change in k.e between time 4 to 6 sec​

Answer 1

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It is given that :-

• Mass of body = 5kg
• Initial Velocity = 0 (since body is at rest)
• Acceleration = 10m/s^2
• ∆KE = ?

Now change in kinetic energy of this body from 4s to 6s will be :

Kinetic energy at 6s - Kinetic energy at 4s.

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Now the velocity of body at 4s :

By first equation of motion :

$v = 0 + 10 \times 4 = 40 \frac{m}{s}$

Now the kinetic energy of body at 4s will be :

$ke = \frac{1}{2} \times m {v}^{2}$

$ke = \frac{1}{2} \times 5 \times 40 \times 40$

$ke = \frac{8000}{2} = 4000j = 4kj$

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Velocity of body at 6s will be :

$v = 0 + 10 \times 6 = 60 \frac{m}{s}$

Now the kinetic energy of body at 6s will be :

$ke = \frac{1}{2} \times m {v}^{2}$

$ke = \frac{1}{2} \times 5 \times 60 \times 60$

$ke = \frac{18000}{2} = 9000j = 9kj$

Therefore the change in kinetic energy will be :

$∆ke = 9kj - 4kj$

$∆ke = 5kj$

Therefore the kinetic energy will increase by 5KJ.

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Answer 2

We have,

Mass of body = 5 kg

u = 0 m/s

a = 10 m/s²

We know,

v = at

⇒v_4 = (10 m/s²)(4 s)

⇒v_4 = 40 m/s

Similarly, v_6 = (10 m/s²)(6 s)

⇒ v_6 = 60 m/s

We know,

∆KE = ½m(v + u)(v - u)

⇒ ∆KE = ½(5 kg)(60 m/s + 40 m/s)(60 m/s - 40 m/s)

⇒ ∆KE = (2.5 kg)(100 m/s)(20 m/s)

⇒ ∆KE = 50 × 100 Nm

⇒ ∆KE = 5000 J = 5 kJ