To solve the problem it is necessary to apply the concepts related to energy conservation. We have a mass of 5Kg that at a distance of 10m has gained 12m / s, with energy losses due to friction.
[tex]KE + W_f = PE[/tex]
[tex]\frac{1}{2}mv^2 +W_f = mgh[/tex]
[tex]W_f = \frac{1}{2}mv^2 -mgh[/tex]
[tex]W_f = \frac{1}{2} 5*12^2-5*9.8*10[/tex]
[tex]W_f = 130J[/tex]
Since there is a loss of energy due to friction, it means that an external force resulting from friction is interacting on the system, which also makes the system open.
The correct answer is D.
There will be some energy loss due to friction countering the non-zero value of external force, which makes the system open. Hence, option (D) is correct.
Given data:
The mass of object is, m = 5 kg.
The falling distance is, h = 10 m.
Speed of object after falling the distance of 10 m is, v = 12 m/s.
This problem can be solved with an approach of concepts related to the energy conservation. We have a mass of 5Kg that at a distance of 10m has gained 12m / s, with energy losses due to friction.
Then as per the conservation of Energy,
[tex]PE = KE + W[/tex]
Here,
PE is the potential energy.
KE is the kinetic energy.
W is the energy losses due to friction.
Solving as,
[tex]PE = KE + W\\\\W=PE - KE\\\\W =mgh-\dfrac{1}{2}mv^{2}\\\\W =(5 \times 9.8 \times 10)-\dfrac{1}{2} \times 5 \times 12^{2}\\\\\\W = 130 \;\rm J[/tex]
So, there will be energy loss of 130 J due to friction, which ultimately means that an external force resulting from friction is interacting on the system. And if there is an interaction of system with surrounding parameters (here it is external force), then the system is open.
Thus, we can conclude that there will be some energy loss due to friction countering the non-zero value of external force, which makes the system open. Hence, option (D) is correct.
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