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garri49 [273]
3 years ago
14

A railroad freight car, mass 15 000 kg, is allowed to coast along a level track at a speed of 2.0 m/s. It collides and couples w

ith a 50 000-kg loaded second car, initially at rest and with brakes released. What percentage of the initial kinetic energy of the 15 000-kg car is preserved in the two-coupled cars after collision
Physics
1 answer:
baherus [9]3 years ago
7 0

Answer:

23.0760769 %

Explanation:

m_1 = Mass of freight car = 15000 kg

m_2 = Mass of second car = 50000 kg

v_1 = Velocity of freight car = 2 m/s

v_2 = Velocity of second car = 0

v = Combined mass velocity

As the linear momentum of the system is conserved we have

m_1v_1+m_2v_2=(m_1+m_2)v\\\Rightarrow v=\dfrac{m_1v_1+m_2v_2}{m_1+m_2}\\\Rightarrow v=\dfrac{15000\times 2+50000\times 0}{15000+50000}\\\Rightarrow v=0.46153\ m/s

The initial kinetic energy

K_i=\dfrac{1}{2}15000\times 2^2\\\Rightarrow K_i=30000\ J

Final kinetic energy

K_f=\dfrac{1}{2}(15000+50000)\times 0.46153^2\\\Rightarrow K_f=6922.82307\ J

The percentage is given by

\dfrac{6922.82307}{30000}\times 100=23.0760769\ \%

The change in percentage of initial kinetic energy is 23.0760769 %

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The force needed to accelerate an elevator upward at a rate of 2 m / s^{2} is 2000 N or 2 kN.

<u>Explanation: </u>

As per Newton's second law of motion, an object's acceleration is directly proportional to the external unbalanced force acting on it and inversely proportional to the mass of the object.

As the object given here is an elevator with mass 1000 kg and the acceleration is given as 2 m / s^{2}, the force needed to accelerate it can be obtained by taking the product of mass and acceleration.

                  \text {Force}=\text {Mass} \times \text {Acceleration}

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