/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 50 A crate of mass \(m\) is release... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 50

A crate of mass \(m\) is released from rest at the top of a stationary ramp of height \(h\), whose length along the curve is \(s\), as shown. Assuming that the ramp is frictionless, in calculating the speed of the crate \(v\) at the bottom of the ramp, one can assume: (A) \(h=v t\) (B) \(m g h=1 / 2 m v^2\) (C) \(s=1 / 2 a t^2\) (D) \(v^2=2 a s\) (E) \(h=1 / 2 g t^2\)

Short Answer

Expert verified
Answer: (B) \(mgh=\frac{1}{2}mv^2\)

Step by step solution

01

- Conservation of Energy

Since there is no friction, mechanical energy is conserved. Initially, the crate has potential energy \(E_p = mgh\). At the bottom of the ramp, the crate has kinetic energy \(E_k = \frac{1}{2}mv^2\). The conservation of energy states that the total energy at the beginning and the end must remain the same. Therefore, we can write the equation as: \(mgh=\frac{1}{2}mv^2\)
02

- Analyzing the options

Now, let's analyze each given option and see whether they match our energy conservation equation or not. (A) \(h=vt\) - This option refers to the height reached by the crate when it's moving at a constant speed. It is unrelated to our problem of finding the speed at the bottom of the ramp. (B) \(mgh=\frac{1}{2}mv^2\) - This option is the same as our energy conservation equation and, therefore, correct. (C) \(s=\frac{1}{2}at^2\) - This kinematic equation refers to the distance covered with a constant acceleration and initial velocity of zero. It is true, but doesn't directly give us the speed of the crate at the bottom of the ramp. (D) \(v^2=2as\) - This kinematic equation relates final speed, initial speed, acceleration, and distance covered. The acceleration isn't constant, so it isn't applicable here. (E) \(h=\frac{1}{2}gt^2\) - This equation refers to the height reached in a vertical motion due to gravity after a time interval \(t\). It is unrelated to our problem. So, the correct expression for calculating the speed of the crate \(v\) at the bottom of the ramp is: (B) \(mgh=\frac{1}{2}mv^2\).

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Key Concepts

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

Potential Energy
Potential energy is a form of stored energy. In this context, when a crate is placed on top of a ramp, it has potential energy due to its position and height from the ground. The equation for potential energy is given by \(E_p = mgh\). Here, \(m\) represents the mass of the crate, \(g\) is the acceleration due to gravity (approximately \(9.8 \text{ m/s}^2\) on Earth), and \(h\) is the height of the ramp.
  • The higher the crate is placed, the more potential energy it possesses.
  • This form of energy is entirely dependent on the position of the object in a gravitational field.
When discussing potential energy in problems involving gravity, remember it's all about height. As the crate moves down the ramp, this energy will convert into another form, which we'll discuss next.
Kinetic Energy
Kinetic energy is the energy of motion. As the crate slides down the ramp and starts moving, its potential energy begins converting into kinetic energy. The formula for kinetic energy is \(E_k = \frac{1}{2}mv^2\), where \(m\) is the mass of the object, and \(v\) is its velocity.
  • As the crate reaches the bottom of the ramp, the potential energy is fully transformed into kinetic energy.
  • This transformation illustrates the principle of conservation of energy, where energy is neither created nor destroyed, only converted from one form to another.
The total energy at the start (potential energy at the top) is equal to the total energy at the end (kinetic energy at the bottom). This concept is fundamental in understanding how objects move and interact in our environment.
Frictionless Ramp
A frictionless ramp is an idealized concept where friction, which typically acts as an opposing force to motion, is absent. In this scenario, the absence of friction means that no energy is lost due to resistance while the crate slides down the ramp.
  • This allows us to use the conservation of energy principle without accounting for energy loss.
  • If there were friction, we would need to compute how much energy is lost to thermal or sound energy.
Without friction, calculations become straightforward since all potential energy converts into kinetic energy as the crate descends. Theoretical scenarios like a frictionless ramp help simplify physics problems and highlight the behavior of energy transformations without external influences.

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Most popular questions from this chapter

A spring obeys Hooke's law. If an amount of work \(W\) is required to stretch the spring a length \(x\) beyond its unstretched length, how much work does it take to stretch it to \(3 x\) ? (A) \(W\) (B) \(3 W\) (C) \(6 W\) (D) \(9 \mathrm{~W}\) (E) \(27 W\)

Two planets, Gamma and Delta, with masses \(m_G\) and \(m_D\) and radii \(R_G\) and \(R_D\) are initially at rest a large distance \(d\) apart from each other. Khan the Conqueror suddenly allows the two planets to approach under their mutual gravitational attraction until they collide. To determine the velocity at which Gamma and Delta collide you could (A) use conservation of energy alone. (B) use conservation of momentum alone. (C) use both conservation of energy and momentum. (D) The problem cannot be solved using energy and momentum conservation. (E) The problem cannot be solved unless you know how long it takes the two planets to collide.

Consider the same setup as in problem 20 , except that now friction exists between the \(15 \mathrm{~kg}\) block and the table, with a coefficient of kinetic friction \(\mu\). Consider the new tension \(T\) in the string and the new acceleration of the blocks, \(a\). One can conclude: (A) \(T\) and \(a\) are both the same as they were in the case without friction. (B) \(T\) and \(a\) are both larger than they were in the case without friction. (C) \(T\) and \(a\) are both smaller than they were in the case without friction. (D) \(T\) is larger than it was in the case without friction, but \(a\) is smaller. (E) \(T\) is smaller than it was in the case without friction, but \(a\) is larger.

An open railroad car is passing at a constant speed \(v_0\) under a water tower, which suddenly dumps a lot of water into the car. The speed of the car will (A) stay the same because of momentum conservation. (B) increase because of momentum conservation. (C) decrease because of momentum conservation. (D) increase because of energy conservation. (E) stay the same because the water is falling in the \(y\)-direction and the car is moving in the \(x\)-direction.

Your dog Astro is nosing a dinner plate of mass \(M\) across a frozen pond at a constant velocity \(v\). There is a coefficient of friction \(\mu\) between the ice and the block. What is the rate of work Astro does on the plate? (A) \(\mu M v^2\) (B) \(\mu M g v\) (C) \(M g v^2 / \mu\) (D) \(\mu M v^2 / g\) (E) \(\mu M g / v\)
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