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2 years ago

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A 2-kg object is moving at 3 m/s. A 4-N force is applied in the direction of motion and if the object has a final velocity 7m/s.

calculate the distance that the object moves

1 answer:

aniked [119]2 years ago

7 0

Intial velocity u=3m/s
final velocity v
2
=u
2
+2as=3
2
+(2×2×5)=29 ⟹v=5.3m/s

KE=
2
1

m(v
2
−u
2
)=
2
1

×2×((5.3)
2
−3
2
)=20J

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Explanation:

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As no external force is acting on the system in the horizontal line of motion. So, the equation will be as follows.

$m_{1}v_{1_{i}} = (m_{1} + m_{2})vi$

where, $m_{1}$ = mass of the projectile

$m_{2}$ = mass of block

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Now, putting the given values into the above formula as follows.

$m_{1}v_{1_{i}} = (m_{1} + m_{2})vi$

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= $\frac{45}{50.075}$

v = 0.898 m/s

Now, equation for energy is as follows.

E = $\frac{1}{2}mv^{2}$

= $\frac{1}{2} \times (75 \times 10^{-3} + 50) \times (600)^{2}$

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Now, energy after the impact will be as follows.

E' = $\frac{1}{2}[75 \times 10^{-3} + 50](0.9)^{2}$

= 20.19 J

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