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liberstina [14]
3 years ago
10

A 80 g particle is moving to the left at 22 m/s . How much net work must be done on the particle to cause it to move to the righ

t at 38 m/s ?
Physics
1 answer:
AysviL [449]3 years ago
8 0

Answer:

W= 38.4 J

Explanation:

Given that

m = 80 g= 0.08 kg

Initial speed ,u= 22 m/s

Final speed ,v= 38 m/s

The change in the kinetic energy of the particle

\Delta KE=\dfrac{1}{2}m(v^2-u^2)

\Delta KE=\dfrac{1}{2}\times 0.08\times (38^2-22^2)\ J

ΔKE= 38.4 J

We know that

Work done by all the forces =Change in the kinetic energy

That is why net work done = 38.4 J.

W= 38.4 J

Therefore the answer will be 38.4 J.

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A 10-g bullet moving horizontally with a speed of 2.0 km/s strikes and passes through a 4.0-kg block moving with a speed of 4.2
SVEN [57.7K]

Answer:

K=512J

Explanation:

Since the surface is frictionless, momentum will be conserved. If the bullet of mass m_1 has an initial velocity v_{1i} and a final velocity v_{1f} and the block of mass m_2 has an initial velocity v_{2i} and a final velocity v_{2f} then the initial and final momentum of the system will be:

p_i=m_1v_{1i}+m_2v_{2i}

p_f=m_1v_{1f}+m_2v_{2f}

Since momentum is conserved, p_i=p_f, which means:

m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}

We know that the block is brought to rest by the collision, which means v_{2f}=0m/s and leaves us with:

m_1v_{1i}+m_2v_{2i}=m_1v_{1f}

which is the same as:

v_{1f}=\frac{m_1v_{1i}+m_2v_{2i}}{m_1}

Considering the direction the bullet moves initially as the positive one, and writing in S.I., this gives us:

v_{1f}=\frac{(0.01kg)(2000m/s)+(4kg)(-4.2m/s)}{0.01kg}=320m/s

So kinetic energy of the bullet as it emerges from the block will be:

K=\frac{mv^2}{2}=\frac{(0.01kg)(320m/s)^2}{2}=512J

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