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Brrunno [24]
3 years ago
10

A sports car accelerates from 0 to 25 meters per second in 4 seconds. What is its acceleration?

Physics
1 answer:
WITCHER [35]3 years ago
3 0

Answer:

6.25 ms²

Explanation:

..................

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A 2kg mass is initially at rest on a frictionless surface. A 45N force acts at an angle of 300
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Answer:

a) The work done by the force is 136.400 joules.

b) The speed of the mass at the end of the 3.5 meters is approximately 11.679 meters per second.

Explanation:

The correct statement is shown below:

<em>A 2-kg mass is initially at reat on a frictionless surface. A 45 N force acts at an angle of 30º to the horizontal for a distance of 3.5 meters:</em>

<em>a)</em><em> Find the work done by the force.</em>

<em>b)</em><em> Find the speed of the mass at the end of the 3.5 meters.</em>

a) Given that a external constant force is acting on the mass on a frictionless surface, the work done by the force (W_{F}), measured in joules, is:

W_{F} = F\cdot \Delta s \cdot \cos \theta (1)

Where:

F - External constant force exerted on the mass, measured in newtons.

\Delta s - Horizontal travelled distance, measured in meters.

\theta - Direction of the external force regarding the horizontal, measured in sexagesimal degrees.

If we know that F = 45\,N, \Delta s = 3.5\,m and \theta = 30^{\circ}, then the work done by the force is:

W_{F} = (45\,N)\cdot (3.5\,m)\cdot \cos 30^{\circ}

W_{F} = 136.400\,J

The work done by the force is 136.400 joules.

b) The final speed of the mass at the end ot the 3.5 meter is calculated by means of the Work-Energy Theorem, which means that the work done on the mass is transformed into a translational kinetic energy. That is:

W_{F} = \frac{1}{2}\cdot m \cdot (v_{2}^{2}-v_{1}^{2}) (2)

Where:

m - Mass, measured in kilograms.

v_{1}, v_{2} - Initial and final speeds of the mass, measured in meters per second.

If we know that W_{F} = 136.400\,J, m = 2\,kg and v_{1} = 0\,\frac{m}{s}, then the final speed of the mass is:

\frac{2\cdot W_{F}}{m} = v_{2}^{2}-v_{1}^{2}

v_{2}^{2} =v_{1}^{2}+\frac{2\cdot W_{F}}{m}

v_{2} =\sqrt{v_{1}^{2}+\frac{2\cdot W_{F}}{m} }

v_{2} =\sqrt{\left(0\,\frac{m}{s} \right)^{2}+\frac{2\cdot (136.400\,J)}{2\,kg} }

v_{2} \approx 11.679\,\frac{m}{s}

The speed of the mass at the end of the 3.5 meters is approximately 11.679 meters per second.

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