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Sagot :
Answer:
[tex]-46.24\ \text{J}[/tex]
Explanation:
[tex]m_1[/tex] = Mass of block at rest = 17 kg
[tex]m_2[/tex] = Mass of moving block = 2.75 kg
[tex]u_1[/tex] = Velocity of block at rest = 0
[tex]u_2[/tex] = Velocity of moving block = 6.25 m/s
v = Velocity of the combined mass
As the momentum in the system is conserved we get
[tex]m_1u_1+m_2u_2=(m_1+m_2)v\\\Rightarrow v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}\\\Rightarrow v=\dfrac{17\times 0+2.75\times 6.25}{17+2.75}\\\Rightarrow v=0.87\ \text{m/s}[/tex]
Change in kinetic energy is given by
[tex]\Delta K=K_f-K_i\\\Rightarrow \Delta K=\dfrac{1}{2}(m_1+m_2)v^2-\dfrac{1}{2}m_2u_2^2\\\Rightarrow \Delta K=\dfrac{1}{2}\times (17+2.75)0.87^2-\dfrac{1}{2}\times 2.75\times 6.25^2\\\Rightarrow \Delta K=-46.24\ \text{J}[/tex]
THe change in kinetic energy in the system is [tex]-46.24\ \text{J}[/tex].
The change in kinetic energy in the system is -46.24 J.
Calculation of the change in the kinetic energy:
Since m1 = Mass of block at rest = 17 kg
m2 = Mass of moving block = 2.75 kg
u1 = Velocity of block at rest = 0
u2 = Velocity of moving block = 6.25 m/s
v = Velocity of the combined mass
So,
momentum in the system should be
[tex]m_1v_1 +m_2v_2 = (m_1 + m_2)v\\\\v = m_1v_1+m_2v_2\div m_1 + m_2\\\\= 17\times 0+2.75\times 6.25 \div 17 + 2.75[/tex]
= 0.87 m/s
Now the change in the kinetic energy should be
[tex]= 1\div 2 (m_1 + m_2)v^2 - 1\div 2 m_2v^2\\\\= 1\div 2\times (17 + 2.75)0.87^2 - 1\div 2\times 2.75\times 6.25^2[/tex]
= -46.24J
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