A 0.40-kg particle moves under the influence of a single conservative force. At point A, where the particle has a speed of 10 m/s, the poten

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A 0.40-kg particle moves under the influence of a single conservative force. At point A, where the particle has a speed of 10 m/s, the potential energy associated with the conservative force is 40 J. As the particle moves from A to B, the force does 25 J of work on the particle. What is the value of the potential energy at point B

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Thu Nguyệt 5 years 2021-08-26T19:32:47+00:00 1 Answers 26 views 0

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  1. Xavia
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    2021-08-26T19:34:06+00:00
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    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:

    K_{A} + U_{A} + W_{A \rightarrow B} = U_{B}

    Where:

    K_{A} – Kinetic energy at point A, measured in joules.

    U_{A} – Potential energy at point A, measured in joules.

    W_{A \rightarrow B} – Work due to conservative force from A to B, measured in joules.

    U_{B} – Potential energy at point B, measured in joules.

    The initial kinetic energy of the particle is:

    K_{A} = \frac{1}{2}\cdot m \cdot v^{2}

    Where:

    m – Mass, measured in kilograms.

    v – Velocity, measured in meters per second.

    If m = 0.4\,kg and v = 10\,\frac{m}{s}, then:

    K_{A} = \frac{1}{2}\cdot (0.4\,kg)\cdot \left(10\,\frac{m}{s} \right)^{2}

    K_{A} = 20\,J

    Finally, the value of the potential energy at point B is:

    U_{B} = 20\,J + 40\,J + 25\,J

    U_{B} = 85\,J

    The value of the potential energy of the particle at point B is 85 joules.

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