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:
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:
a) Find the work done by the force.
b) Find the speed of the mass at the end of the 3.5 meters.
a) Given that a external constant force is acting on the mass on a frictionless surface, the work done by the force ([tex]W_{F}[/tex]), measured in joules, is:
[tex]W_{F} = F\cdot \Delta s \cdot \cos \theta[/tex] (1)
Where:
[tex]F[/tex] - External constant force exerted on the mass, measured in newtons.
[tex]\Delta s[/tex] - Horizontal travelled distance, measured in meters.
[tex]\theta[/tex] - Direction of the external force regarding the horizontal, measured in sexagesimal degrees.
If we know that [tex]F = 45\,N[/tex], [tex]\Delta s = 3.5\,m[/tex] and [tex]\theta = 30^{\circ}[/tex], then the work done by the force is:
[tex]W_{F} = (45\,N)\cdot (3.5\,m)\cdot \cos 30^{\circ}[/tex]
[tex]W_{F} = 136.400\,J[/tex]
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:
[tex]W_{F} = \frac{1}{2}\cdot m \cdot (v_{2}^{2}-v_{1}^{2})[/tex] (2)
Where:
[tex]m[/tex] - Mass, measured in kilograms.
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final speeds of the mass, measured in meters per second.
If we know that [tex]W_{F} = 136.400\,J[/tex], [tex]m = 2\,kg[/tex] and [tex]v_{1} = 0\,\frac{m}{s}[/tex], then the final speed of the mass is:
[tex]\frac{2\cdot W_{F}}{m} = v_{2}^{2}-v_{1}^{2}[/tex]
[tex]v_{2}^{2} =v_{1}^{2}+\frac{2\cdot W_{F}}{m}[/tex]
[tex]v_{2} =\sqrt{v_{1}^{2}+\frac{2\cdot W_{F}}{m} }[/tex]
[tex]v_{2} =\sqrt{\left(0\,\frac{m}{s} \right)^{2}+\frac{2\cdot (136.400\,J)}{2\,kg} }[/tex]
[tex]v_{2} \approx 11.679\,\frac{m}{s}[/tex]
The speed of the mass at the end of the 3.5 meters is approximately 11.679 meters per second.
