Question

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

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

Answer 1

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.

Step-by-step 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 (
`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) = (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\,(m)/(s)`, then the final speed of the mass is:

`(2\cdot W_(F))/(m) = v_(2)^(2)-v_(1)^(2)`

`v_(2)^(2) =v_(1)^(2)+(2\cdot W_(F))/(m)`

`v_(2) =\sqrt{v_(1)^(2)+(2\cdot W_(F))/(m) }`

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

`v_(2) \approx 11.679\,(m)/(s)`

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