A physics professor is pushed up a ramp inclined upward at 30.0° above the horizontal as she sits in her desk chair, which slides on frictionless rollers. the combined mass of the professor and chair is 85.0 kg. she is pushed 2.50 m along the incline by a group of students who together exert a constant horizontal force of 600 n. the professor’s speed at the bottom of the ramp is 2.00 m>s. use the work–energy theorem to find her speed at the top of the ramp.

[tex]F_{net}d = \frac{1}{2}mv^{2} - \frac{1}{2}mu^{2}\\\frac{2F_{net}d }{m} = v^{2} -u^{2}\\ v^{2} =\frac{2F_{net}d }{m} +u^{2}\\ v^{2} =\frac{2(600cos30 - 85\times 9.8 \times sin30) \times 2.5 }{85} +2.00^{2}\\ v^{2} = 10.066\\v = 3..17m/s[/tex]

Lanuel Lanuel · Answer 2 · 2022-02-25T12:52:06+01:00

The professor’s speed at the top of the ramp is equal to 3.17 m/s.

Given the following data:

Angle of inclination = 30.0°

Mass of professor and chair = 85.0 kg.

Displacement = 2.50 meters.

Force = 600 Newton.

Professor’s speed = 2 m/s.

Scientific data:

Acceleration due to gravity = 9.8 [tex]m/s^2[/tex].

To determine the professor’s speed at the top of the ramp, we would apply the work-energy theorem:

The application of work-energy theorem.

According to the work-energy theorem, the work done by an object is equal to the change in kinetic energy possessed by the object.

Mathematically, this is given by this expression:

[tex]W = \Delta K.E\\\\Fd = \frac{1}{2} mv^2 - \frac{1}{2} mu^2\\\\\frac{2Fd}{m} =v^2-u^2\\\\v^2 = \frac{2Fd}{m} +u^2\\\\v = \sqrt{\frac{2Fd}{m} +u^2}[/tex]

Substituting the given parameters into the formula, we have;

[tex]v = \sqrt{\frac{2\times [600cos30-85 \times 9.8 \times sin30] \times 2.5}{85} +2^2}\\\\v = \sqrt{\frac{2\times [519.62-416.5] \times 2.5}{85} +4}\\\\v=\sqrt{6.0659 +4} \\\\v=\sqrt{10.0659}[/tex]

Speed, v = 3.17 m/s.

Read more on work done here: brainly.com/question/22599382