Answer :
Answer:
The speed of object B is 2 times of speed of object A.
Explanation:
Given that,
Kinetic energy of object A = 25 J
Mass of object B [tex]m_{B}=\dfrac{1}{4}m_{A}[/tex]
Work done = -20 J
We need to calculate the factor of the speed of the object
Suppose the final kinetic energy is same for both object.
[tex]K.E_{f_{A}}=K.E_{f_{B}}[/tex]
[tex]\dfrac{1}{2}m_{A}v_{A}^2=\dfrac{1}{2}m_{B}v_{B}^2[/tex]
Put the value into the formula
[tex]\dfrac{1}{2}m_{A}\times v_{A}^2=\dfrac{1}{2}\times\dfrac{1}{4}m_{A}\times v_{B}^2[/tex]
[tex]\dfrac{v_{A}}{v_{B}}=\dfrac{1}{2}[/tex]
[tex]v_{A}=\dfrac{1}{2}v_{B}[/tex]
[tex]v_{B}=2v_{A}[/tex]
Hence, The speed of object B is 2 times of speed of object A.
The factor at which the speed of each object change if the total work is -20 J is done on each is [tex]\mathbf{V_A' = 0.45 \ V_A }[/tex] and [tex]\mathbf{V_B' = 0.45 \ V_B }[/tex]
Kinetic energy is an energy that is said to be in motion. This occurs by the work required to move a mass of an object from a state of rest to a specified
velocity.
From the given information:
Let consider the mass of B to be [tex]\mathbf{m_B = \dfrac{m_A}{4}}[/tex]
The kinetic energy of object A = kinetic energy of object B.
∴
[tex]\mathbf{E_{KA}=E_{KB}}[/tex] = 25 J
[tex]\mathbf{\dfrac{1}{2}m_AV_A^2 =\dfrac{1}{2}m_BV_B^2}[/tex]
[tex]\mathbf{\dfrac{V_B}{V_A} = (\dfrac{m_A}{m_B})^{1/2} }[/tex]
Recall that: [tex]\mathbf{m_B = \dfrac{m_A}{4}}[/tex]
∴
[tex]\mathbf{\dfrac{V_B}{V_A} = 2}[/tex]
[tex]\mathbf{V_B = 2V_A}[/tex]
It implies that object B is twice as faster than object A
However, suppose the kinetic energy of A after an amount of work done
W = -20 J
It implies that [tex]\mathbf{E_{KA}' - E_{KA} = W}[/tex]
[tex]\mathbf{E_{KA}' = W+E_{KA}}[/tex]
[tex]\mathbf{E_{KA}' = -20+25}[/tex]
[tex]\mathbf{E_{KA}' =5 \ J}[/tex]
[tex]\mathbf{E_{KA}' =E_{KB}' = 5 \ J}[/tex]
∴
The factor by which the speed(V) of each object changes can be computed as:
[tex]\mathbf{\dfrac{E_{KA}' }{ E_{kA}} = \dfrac{5}{25}}[/tex]
[tex]\mathbf{\dfrac{\dfrac{1}{2}m_A V_A^2' }{ \dfrac{1}{2}mV_A^2} = \dfrac{1}{5}}[/tex]
[tex]\mathbf{V_A' = \dfrac{1}{\sqrt{5}}V_A}[/tex]
[tex]\mathbf{V_A' = 0.45 \ V_A }[/tex]
Similarly; [tex]\mathbf{V_B' = 0.45 \ V_B }[/tex]
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