Question

A 0.54 kg particle has a speed of8.0 m/s at point A and kineticenergy of 7.5 J at point B.

Answer 1

Answer:

a) The translational kinetic energy of the particle at point A is 17.28 joules.

b) The speed of the particle at point B is approximately 5.270 meters per second.

c) The total work done on the particle as it moves from A to B is - 9.78 joules.

Explanation:

Let be this particle a conservative system, that is, that non-conservative forces (i.e. friction, viscosity) are negligible.

a) The translational kinetic energy of the particle ($K$), measured in joules, is determined by the following formula:

$K = \frac{1}{2}\cdot m \cdot v^{2}$ (1)

Where:

$m$ - Mass, measured in kilograms.

$v$ - Speed, measured in meters per second.

If we know that $m = 0.54\,kg$ and $v = 8\,\frac{m}{s}$, the translational kinetic energy at point A is:

$K = \frac{1}{2}\cdot (0.54\,kg)\cdot \left(8\,\frac{m}{s} \right)^{2}$

$K = 17.28\,J$

The translational kinetic energy of the particle at point A is 17.28 joules.

b) The speed of the particle is clear in (1):

$v = \sqrt{\frac{2\cdot K}{m} }$

If we know that $K = 7.5\,J$ and $m = 0.54\,kg$, then the speed of the particle at point B:

$v = \sqrt{\frac{2\cdot (7.5\,J)}{0.54\,kg} }$

$v\approx 5.270\,\frac{m}{s}$

The speed of the particle at point B is approximately 5.270 meters per second.

c) According to the Work-Energy Theorem, the total work done on the particle as it moves from A to B ($W_{A\rightarrow B}$), measured in joules, is equal to the change in the translational kinetic energy of the particle. That is:

$W_{A\rightarrow B} = K_{B}-K_{A}$ (2)

If we know that $K_{A} = 17.28\,J$ and $K_{B} = 7.50\,J$, then the change in the translational kinetic energy of the particle is:

$W_{A\rightarrow B} = 7.50\,J-17.28\,J$

$W_{A\rightarrow B} = -9.78\,J$

The total work done on the particle as it moves from A to B is - 9.78 joules.