Respuesta :
Answer:
The value of the potential energy of the particle at point B is 85 joules.
Explanation:
According to the Principle of Energy Conservation, the energy cannot be created nor destroyed, only transformed. The particle at point A has kinetic and potential energy and receives a work due to an external conservative force (Work-Energy Theorem), whose sum is equal to potential energy at point B. Mathematically speaking, the expression that describes the phenomenon is:
[tex]K_{A} + U_{A} + W_{A \rightarrow B} = U_{B}[/tex]
Where:
[tex]K_{A}[/tex] - Kinetic energy at point A, measured in joules.
[tex]U_{A}[/tex] - Potential energy at point A, measured in joules.
[tex]W_{A \rightarrow B}[/tex] - Work due to conservative force from A to B, measured in joules.
[tex]U_{B}[/tex] - Potential energy at point B, measured in joules.
The initial kinetic energy of the particle is:
[tex]K_{A} = \frac{1}{2}\cdot m \cdot v^{2}[/tex]
Where:
[tex]m[/tex] - Mass, measured in kilograms.
[tex]v[/tex] - Velocity, measured in meters per second.
If [tex]m = 0.4\,kg[/tex] and [tex]v = 10\,\frac{m}{s}[/tex], then:
[tex]K_{A} = \frac{1}{2}\cdot (0.4\,kg)\cdot \left(10\,\frac{m}{s} \right)^{2}[/tex]
[tex]K_{A} = 20\,J[/tex]
Finally, the value of the potential energy at point B is:
[tex]U_{B} = 20\,J + 40\,J + 25\,J[/tex]
[tex]U_{B} = 85\,J[/tex]
The value of the potential energy of the particle at point B is 85 joules.
The potential energy of the particle at point B is 85 J.
Given to us:
Mass of the particle, [tex]m=0.40\ kg[/tex]
velocity of the particle, [tex]v= 10\ m/s[/tex]
potential energy of the particle, [tex]PE= 40\ J[/tex]
Workdone from pt. A to B, [tex]WD_{(A\rightarrow B)} = 25\ J[/tex]
Calculating the kinetic energy of the particle,
[tex]\begin{aligned}KE&= \frac{1}{2}mv^2 \\\\&=\frac{1}{2}\times0.40\times (10)^2\\\\&=20 J\\\end{aligned}[/tex]
According to the Principle of Energy Conservation,
The energy cannot be created nor be destroyed, it can only be transformed from one form to another.Therefore,
Also,
Total Energy at point A ,
[tex]\begin{aligned}(TE)_A &= Kinetic\ Energy+ Potential\ Energy+ Internal Energy+ Workdone\\ &=KE_A+ PE_A+UE_A+ WD_{(0\rightarrow A)}\\&=20+40+0+0\\&=60\ J\end{aligned}[/tex]
Total Energy at point B,
[tex]\begin{aligned}(TE)_B &= Kinetic\ Energy+ Potential\ Energy+ Internal Energy+ Workdone\\ &=KE_B+ PE_B+UE_B+ WD_{(A\rightarrow B)}\end{aligned}[/tex]
As the total energy is conserved from point A to B and also an external work is done on the particle. we can write the above equation as,
[tex]\begin{aligned} TE_B&=KE_B+ PE_B+UE_B+ WD_{(A\rightarrow B)}\\&=(KE_B+ PE_B+UE_B)+ WD_{(A\rightarrow B)}\\&= TE_A+ WD_{(A\rightarrow B)}\\&=60+25\\&=85\ J\end{aligned}[/tex]
Therefore, the total energy for the particle at point B is 85 J but as the particle is not moving neither work is done at point B, the total energy of the particle is the potential energy of the particle.
Hence, the potential energy of the particle at point B is 85 J.
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