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Chapter 6: Problem 80

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 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.

Short Answer

Expert verified
The professor's speed at the top of the ramp is approximately 3.18 m/s.

Step by step solution

01

Calculate the Initial Kinetic Energy

Using the formula for kinetic energy (KE), calculate the initial kinetic energy at the bottom of the ramp. The initial speed at the bottom of the ramp is given as 2.00 m/s. The formula for kinetic energy is:\[ \text{KE}_i = \frac{1}{2} mv_i^2 \]where \(m = 85.0\, \text{kg}\) and \(v_i = 2.00\, \text{m/s}\). Substitute the values:\[ \text{KE}_i = \frac{1}{2} \times 85.0 \times (2.00)^2 = 170.0\, \text{J} \]
02

Calculate the Work Done by the Students

The students exert a constant horizontal force of 600 N while moving the professor 2.50 m along the incline. Calculate the work done (W) using the formula:\[ W = Fd \cos(\theta) \]where \(F = 600\, \text{N}\), \(d = 2.50\, \text{m}\), and \(\theta = 30.0^\circ\). Substitute the values:\[ W = 600 \times 2.50 \times \cos(30.0^\circ) \approx 1299.04\, \text{J} \]
03

Calculate the Gravitational Potential Energy Increase

The professor is moved up a height \(h\) along the incline. The height can be determined using the relation \( h = d \sin(\theta) \). The increase in gravitational potential energy (GPE) is given by:\[ \text{GPE} = mgh \]Substitute the values to find \(h\):\[ h = 2.50 \times \sin(30.0^\circ) = 1.25\, \text{m} \] Now calculate GPE:\[ \text{GPE} = 85.0 \times 9.81 \times 1.25 \approx 1040.13\, \text{J} \]
04

Apply the Work-Energy Theorem

According to the work-energy theorem, the work done by the students plus the initial kinetic energy is equal to the final kinetic energy at the top of the ramp plus the gravitational potential energy:\[ W + \text{KE}_i = \text{KE}_f + \text{GPE} \]Substitute the known values:\[ 1299.04 + 170.0 = \text{KE}_f + 1040.13 \]Solve for \(\text{KE}_f\):\[ \text{KE}_f = 1299.04 + 170.0 - 1040.13 = 428.91\, \text{J} \]
05

Calculate the Final Speed

Now, use the formula for kinetic energy to find the final speed \(v_f\) at the top of the ramp:\[ \text{KE}_f = \frac{1}{2} mv_f^2 \]Solve for \(v_f\):\[ 428.91 = \frac{1}{2} \times 85.0 \times v_f^2 \]\[ v_f^2 = \frac{428.91 \times 2}{85.0} = 10.097 \]\[ v_f = \sqrt{10.097} \approx 3.18\, \text{m/s} \]

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy possessed by an object due to its motion. It's a fundamental concept in physics that helps us understand how objects change speed and move. The formula to calculate kinetic energy (KE) is \( KE = \frac{1}{2} mv^2 \), where \( m \) represents mass and \( v \) is velocity.

In the original exercise, the physics professor's speed when starting to move up the ramp was 2.00 m/s, and, together with the chair, had a mass of 85.0 kg.

By substituting these values into the formula, we calculated the initial kinetic energy at the bottom of the ramp to be 170.0 J. Kinetic energy increases with the square of the velocity, which means even a slight increase in speed significantly increases kinetic energy.

Understanding this allows us to see the effects of forces on moving objects.
Gravitational Potential Energy
Gravitational potential energy (GPE) represents the energy stored in an object as it is situated in a gravitational field. It is specifically based on the object's height relative to a reference point, such as the ground.

The formula used to determine gravitational potential energy is \( GPE = mgh \), where \( m \) is mass, \( g \) is acceleration due to gravity, and \( h \) is height. When the professor was moved up the incline, she was lifted to a new height, thereby gaining potential energy.

In this case, the ramp's incline helped us calculate the real height gained using trigonometry. The ramp's height gain was found to be 1.25 m, which translates to a potential energy increase of approximately 1040.13 J.
  • This increase happens because as the professor rises, the gravitational pull stores energy in her system.
  • Gaining height increases gravitational potential energy, while losing height decreases it.
Inclined Plane
An inclined plane is a flat surface tilted at an angle, which helps lift objects by using less force over a longer distance. It exemplifies how simple machines make work easier by redistributing force.

In the scenario with the physics professor, the ramp was inclined at a 30.0掳 angle. This means the students needed to exert force mainly horizontally while gravity worked against them vertically. Calculating movements on an inclined plane often involves using trigonometry. Here, it helped compute both the work done along the slope and the increase in height.
  • Inclined planes reduce the force needed to move objects upwards, compared to lifting straight up.
  • Using an inclined plane can be an efficient way to overcome vertical obstacles with steady effort.
Constant Force
A constant force functions as a steady influence that doesn't change in magnitude or direction over time. When applied to an object, this consistent energy can change the object's motion in a predictable pattern.

In this physics problem, students were able to apply a constant force of 600 N to push the professor up the ramp. This regular application of force, even though horizontal, contributed to her kinetic energy as she ascended.

By calculating the work done, \( W = Fd\cos(\theta) \), we integrated the strength of the push (600 N), the distance moved (2.50 m), and the incline's angle (30.0掳), resulting in 1299.04 J of work done.
  • The concept of constant force illustrates how regular energy input can result in a calculated shift in momentum or energy.
  • Constant forces often simplify calculations in physics by allowing steady comparisons.

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Most popular questions from this chapter

A force in the \(+x\)-direction with magnitude \(F(x) = 18.0 \, \mathrm{N} - (0.530 \, \mathrm{N/m})x\) is applied to a 6.00-kg box that is sitting on the horizontal, frictionless surface of a frozen lake. \(F(x)\) is the only horizontal force on the box. If the box is initially at rest at \(x = 0\), what is its speed after it has traveled 14.0 m?

A mass \(m\) slides down a smooth inclined plane from an initial vertical height \(h\), making an angle \(\alpha\) with the horizontal. (a) The work done by a force is the sum of the work done by the components of the force. Consider the components of gravity parallel and perpendicular to the surface of the plane. Calculate the work done on the mass by each of the components, and use these results to show that the work done by gravity is exactly the same as if the mass had fallen straight down through the air from a height \(h\). (b) Use the work\(-\)energy theorem to prove that the speed of the mass at the bottom of the incline is the same as if the mass had been dropped from height \(h\), independent of the angle \(\alpha\) of the incline. Explain how this speed can be independent of the slope angle. (c) Use the results of part (b) to find the speed of a rock that slides down an icy frictionless hill, starting from rest 15.0 m above the bottom.

(a) Suppose you cut a massless ideal spring in half. If the full spring had a force constant k, what is the force constant of each half, in terms of \(k\)? (\(Hint\): Think of the original spring as two equal halves, each producing the same force as the entire spring. Do you see why the forces must be equal?) (b) If you cut the spring into three equal segments instead, what is the force constant of each one, in terms of \(k\)?

Your job is to lift 30-kg crates a vertical distance of 0.90 m from the ground onto the bed of a truck. How many crates would you have to load onto the truck in 1 minute (a) for the average power output you use to lift the crates to equal 0.50 hp; (b) for an average power output of 100 W?

A 4.00-kg block of ice is placed against a horizontal spring that has force constant \(k\) = 200 N/m and is compressed 0.025 m. The spring is released and accelerates the block along a horizontal surface. Ignore friction and the mass of the spring. (a) Calculate the work done on the block by the spring during the motion of the block from its initial position to where the spring has returned to its uncompressed length. (b) What is the speed of the block after it leaves the spring?
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