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A physics professor is pushed up a ramp inclined upward at \(30.0^{\circ}\) above the horizontal as she sits in her desk chair, which slides on frictionless rollers. The combined mass of the professor and chair is \(85.0 \mathrm{~kg} .\) She is pushed \(2.50 \mathrm{~m}\) along the incline by a group of students who together exert a constant horizontal force of \(600 \mathrm{~N}\). The professor's speed at the bottom of the ramp is \(2.00 \mathrm{~m} / \mathrm{s}\). Use the work-energy theorem to find her speed at the top of the ramp.

Short Answer

Expert verified
Professor's speed at the top of the ramp can be calculated by first finding the initial kinetic energy at the bottom of the ramp, then determining the work done by the students, applying the work-energy theorem, and finally using the final kinetic energy to find the speed at the top of the ramp. The method to arrive the result includes multiple steps involving kinematics and calculations of work and energy.

Step by step solution

01

Calculate Initial Kinetic Energy

Firstly, calculate the initial kinetic energy at the bottom of the ramp using formula \(K.E. = \frac{1}{2}mv^2\). Here m = 85.0 kg (combined mass of the professor and chair) and v = 2.00 m/s (speed of professor at the bottom of the ramp). Thus, \(K.E. = \frac{1}{2} \times 85.0 \times (2.00)^2\) Joules.
02

Calculate Work done

Next, calculate the work done by the students. Work done is force (F) multiplied by the distance (d) times the cosine of the angle (θ), \(W = Fdcosθ\). Here, F = 600 N (horizontal force exerted), d = 2.50 m (distance along the incline), and θ = 180 - 30 (angle between force and displacement is supplementary to the angle of incline). Therefore, we get \(W = 600 \times 2.50 \times \cos(150)\) Joules.
03

Apply Work-Energy Theorem

Apply work-energy theorem which states that work done is equal to the change in kinetic energy. In this scenario, ΔK.E. = K.E._final - K.E._initial. Hence, K.E._final = W + K.E._initial.
04

Find the Final Speed

Finally, use the kinetic energy formula to find the final speed at the top. The final kinetic energy is calculated in step 3, and mass m = 85.0 kg. Use the formula to get \(v_f = \sqrt{2 \times \frac{K.E._final}{m}}\) m/s.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Energy
Understanding kinetic energy is crucial when analyzing the movement of objects. It's the energy an object possesses due to its motion; think of it as the work needed to accelerate an object to a certain velocity. The kinetic energy (\textbf{K.E.}) of an object is given by the formula: \[ K.E. = \frac{1}{2}mv^2 \] where \(m\) represents the mass of the object and \(v\) is its velocity.

The initial kinetic energy in our exercise is calculated by plugging in the professor and chair's combined mass and the chair's initial speed. Because the professor is initially moving, we know that some amount of work has already been done to get her to that speed. The calculation of kinetic energy provides a numerical value for that work.
Work Done by Force
Work done by a force is another fundamental concept in physics, telling us how much energy has been transferred by a force moving an object over a distance. The general formula is \[ W = Fd\cos(\theta) \] 'F' is the magnitude of the force, 'd' is the distance over which the force is applied, and \(\theta\) is the angle between the direction of the force and the direction of displacement.

In the case of the professor being pushed up an inclined ramp, the force applied by the students is horizontal, but the professor moves along the incline. Therefore, we take into account the angle between the force and the direction of movement, which affects the amount of work that directly goes into moving the professor up the incline. Through trigonometry, we relate the applied force to the actual work done against gravity, allowing us to calculate the energy transferred to the professor and chair.
Inclined Plane Physics
Inclined planes are surfaces tilted at an angle to horizontal, leading to interesting applications of Newton's second law of motion. They are great examples of how forces can be resolved into components that act parallel and perpendicular to the surface.

Inclined plane problems often involve the force of gravity, friction forces, normal forces, and applied forces. The significant aspect to remember here is the effect of the plane angle on the various component forces which ultimately influences the work done on the object. In this exercise, the force applied is horizontal, but the point of interest is the movement along the slope, so the angle used in the work done calculation is based on the incline of the ramp.
Conservation of Energy
The concept of the conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. In our physics scenario, this principle ensures that the total energy at the beginning (initial kinetic energy) and the energy at the end (final kinetic energy plus the work done by the students) will be equal.

When the students push the professor up the ramp, they are doing work against gravity. This work is converted into additional kinetic energy if there's no friction. The work-energy theorem connects this principle to our problem, stating that the work done on an object is equivalent to the change in the object’s kinetic energy. So, the initial kinetic energy plus the work done by the students' force gives us the final kinetic energy from which we can deduce the professor's speed at the top of the ramp.

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Most popular questions from this chapter

Using a cable with a tension of \(1350 \mathrm{~N}\), a tow truck pulls a car \(5.00 \mathrm{~km}\) along a horizontal roadway. (a) How much work does the cable do on the car if it pulls horizontally? If it pulls at \(35.0^{\circ}\) above the horizontal? (b) How much work does the cable do on the tow truck in both cases of part (a)? (c) How much work does gravity do on the car in part (a)?

A \(75.0 \mathrm{~kg}\) painter climbs a ladder that is \(2.75 \mathrm{~m}\) long and leans against a vertical wall. The ladder makes a \(30.0^{\circ}\) angle with the wall. (a) How much work does gravity do on the painter? (b) Does the answer to part (a) depend on whether the painter climbs at constant speed or accelerates up the ladder?

Proton Bombardment. A proton with mass \(1.67 \times 10^{-27} \mathrm{~kg}\) is propelled at an initial speed of \(3.00 \times 10^{5} \mathrm{~m} / \mathrm{s}\) directly toward a uranium nucleus \(5.00 \mathrm{~m}\) away. The proton is repelled by the uranium nucleus with a force of magnitude \(F=\alpha / x^{2},\) where \(x\) is the separation between the two objects and \(\alpha=2.12 \times 10^{-26} \mathrm{~N} \cdot \mathrm{m}^{2}\). Assume that the uranium nucleus remains at rest. (a) What is the speed of the proton when it is \(8.00 \times 10^{-10} \mathrm{~m}\) from the uranium nucleus? (b) As the proton approaches the uranium nucleus, the repulsive force slows down the proton until it comes momentarily to rest, after which the proton moves away from the uranium nucleus. How close to the uranium nucleus does the proton get? (c) What is the speed of the proton when it is again \(5.00 \mathrm{~m}\) away from the uranium nucleus?

On an essentially frictionless, horizontal ice rink, a skater moving at \(3.0 \mathrm{~m} / \mathrm{s}\) encounters a rough patch that reduces her speed to \(1.65 \mathrm{~m} / \mathrm{s}\) due to a friction force that is \(25 \%\) of her weight. Use the work-energy theorem to find the length of this rough patch.

A 12-pack of Omni-Cola (mass \(4.30 \mathrm{~kg}\) ) is initially at rest on a horizontal floor. It is then pushed in a straight line for \(1.20 \mathrm{~m}\) by a trained dog that exerts a horizontal force with magnitude \(36.0 \mathrm{~N}\). Use the work-energy theorem to find the final speed of the 12 -pack if (a) there is no friction between the 12 -pack and the floor, and (b) the coefficient of kinetic friction between the 12 -pack and the floor is 0.30 .

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