Question

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

Answer 1

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.