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Prove that if the block is released from rest at point \(B\) of a smooth path of arbitrary shape, the speed it attains when it reaches point \(A\) is equal to the speed it attains when it falls freely through a distance \(h ;\) i.e., \(v=\sqrt{2 g h}.\)

Short Answer

Expert verified
The block will attain a speed of \(v = \sqrt{2gh}\) when it reaches point \(A\), same as it would have if it had fallen freely through the same height \(h\). This is a consequence of the conservation of mechanical energy principle.

Step by step solution

01

Identify Relevant Energy Forms

The block starts with potential energy at point \(B\) and ends with kinetic energy at point \(A\). At the beginning, its velocity is zero, so its kinetic energy is also zero. The potential energy is given by \(U = mgh\) where \(m\) is the mass of the block, \(g\) is the acceleration due to gravity, and \(h\) is the height of point \(B\). At point \(A\), the block has a velocity \(v\), so its kinetic energy is \(K = 0.5mv^2\). Its potential energy at \(A\) is zero because we consider \(A\) to be our reference ground level.
02

Apply Conservation of Energy

According to the principle of conservation of energy, the total energy at the starting point should be equal to the total energy at the ending point. This implies that the potential energy at point \(B\) is equal to the kinetic energy at point \(A\) because there are no non-conservative forces (like friction) at play. Mathematically, this can be written as \(mgh= \frac{1}{2} m v^2\). In this equation, we can see that the mass \(m\) comes on both sides, so it can be divided out.
03

Solve for Velocity

After canceling the mass in the equation \(mgh= \frac{1}{2} m v^2\), the equation becomes \(2gh = v^2\). Solving for \(v\) by taking the square root of both sides, we get \(v = \sqrt{2gh}\). Which is the speed that the block attains when it reaches point \(A\).

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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 that an object has because of its motion. When a block or any object is moving, it possesses kinetic energy. The amount of kinetic energy an object has depends on two things: its mass and its speed. The mathematical formula to calculate kinetic energy is \(K = \frac{1}{2}mv^2\), where \(m\) represents the mass of the object and \(v\) is its velocity.

This concept is vital in understanding energy conservation experiments like the one given in the exercise. At point \(A\), the block converts its potential energy into kinetic energy, resulting in the formula \(K = \frac{1}{2}mv^2\). Thus, kinetic energy helps in determining how fast an object will move once it reaches the bottom of its trajectory. It's fascinating because even if an object's mass changes, its velocity calculations remain unaffected when using energy calculations because the mass cancels out in equations like conservation of energy.
Potential Energy
Potential energy, on the other hand, is the energy stored in an object because of its position or state. When you lift a block to a certain height, it gains potential energy because it has the potential to convert that energy into motion.

In this exercise, when the block is at point \(B\), it possesses potential energy calculated as \(U = mgh\), where \(m\) is mass, \(g\) is acceleration due to gravity, and \(h\) is the height at which the block is placed. As the block moves downwards to point \(A\), this stored potential energy decreases since it’s being converted into kinetic energy.

It's interesting how potential energy gives us insights into how much work might be done by the object as it moves. So, whenever you see an object positioned at a height, remember it has the potential energy ready to be converted into other energy forms.
Energy Transformation
Energy transformation is a key idea illustrating how energy shifts from one form to another. In almost every physical scenario, energy doesn’t just appear or disappear; it transforms.

In our exercise, energy transformation is observed when the block moves from a higher position at point \(B\) to a lower position at point \(A\). Initially, all energy is potential, stored due to the height of the block. As it descends, this potential energy gets converted into kinetic energy, propelling the block forward with a specific velocity, \(v = \sqrt{2gh}\). This equation represents the energy transformation from potential to kinetic energy perfectly and shows how height and gravitational forces work together in this conversion process.

This transformation underscores the beauty of the conservation of energy principle. It assures us that total energy stays constant, only transforming from one type to another without any loss. Hence, understanding energy transformation helps us grasp how energies interact in the universe efficiently and predictably.

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Most popular questions from this chapter

If the 50 -kg crate starts from rest and achieves a velocity of \(v=4 \mathrm{m} / \mathrm{s}\) when it travels a distance of \(5 \mathrm{m}\) to the right, determine the magnitude of force \(\mathbf{P}\) acting on the crate. The coefficient of kinetic friction between the crate and the ground is \(\mu_{k}=0.3.\)

Block \(A\) has a weight of 8 lb and block \(B\) has a weight of 6 lb. They rest on a surface for which the coefficient of kinetic friction is \(\mu_{k}=0.2 .\) If the spring has a stiffness of \(k=20\) lb/ft, and it is compressed \(0.2 \mathrm{ft}\), determine the acceleration of each block just after they are released.

Rod \(O A\) rotates counterclockwise with a constant angular velocity of \(\dot{\theta}=5 \mathrm{rad} / \mathrm{s} .\) The double collar \(B\) is pinconnected together such that one collar slides over the rotating rod and the other slides over the horizontal curved rod, of which the shape is described by the equation \(r=1.5(2-\cos \theta)\) ft. If both collars weigh 0.75 lb, determine the normal force which the curved rod exerts on one collar at the instant \(\theta=120^{\circ} .\) Neglect friction.

The 10 -lb block has a speed of 4 ft / s when the force of \(F=\left(8 t^{2}\right)\) lb is applied. Determine the velocity of the block when \(t=2\) s. The coefficient of kinetic friction at the surface is \(\mu_{k}=0.2\).

A girl, having a mass of 15 kg, sits motionless relative to the surface of a horizontal platform at a distance of \(r=5 \mathrm{m}\) from the platform's center. If the angular motion of the platform is slowly increased so that the girl's tangential component of acceleration can be neglected, determine the maximum speed which the girl will have before she begins to slip off the platform. The coefficient of static friction between the girl and the platform is \(\mu=0.2\).

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