/*! 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 132 A block of metal weighing \(2 \m... [FREE SOLUTION] | 91Ó°ÊÓ

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A block of metal weighing \(2 \mathrm{~kg}\) is resting on a frictionless plane. It is struck by a jet releasing water at a rate of 1 \(\mathrm{kg} / \mathrm{s}\) and at a speed of \(5 \mathrm{~m} / \mathrm{s}\). The initial acceleration of the block is a. \(\frac{5}{3} \mathrm{~m} / \mathrm{s}^{2}\) b. \(\frac{25}{4} \mathrm{~m} / \mathrm{s}^{2}\) c. \(\frac{25}{8} \mathrm{~m} / \mathrm{s}^{2}\) d. \(\frac{5}{2} \mathrm{~m} / \mathrm{s}^{2}\)

Short Answer

Expert verified
The initial acceleration of the block is \(\frac{5}{2}\, \mathrm{m/s^2}\), which is option d.

Step by step solution

01

Understand the Problem

We have a block of metal with a mass of \(2\, \mathrm{kg}\) resting on a frictionless plane. It is struck by a jet that releases water at a rate of \(1\, \mathrm{kg/s}\) with a speed of \(5\, \mathrm{m/s}\). We need to find the initial acceleration of the block.
02

Apply the Principle of Momentum

Since the jet is striking the block with water, we can use the principle of momentum. The force exerted by the water jet on the block is equal to the rate of change of momentum of the water.
03

Calculate Change in Momentum

The momentum change of the water per second is given by the product of the rate of water (1 \, \mathrm{kg/s}) and the speed of the water (5\, \mathrm{m/s}). Thus, the force \(F\) exerted on the block is: \[ F = \text{mass rate} \times \text{velocity} = 1\, \mathrm{kg/s} \times 5\, \mathrm{m/s} = 5\, \mathrm{N} \].
04

Use Newton's Second Law

Using Newton's second law of motion, \(F = ma\), where \(m\) is the mass of the block and \(a\) is its acceleration. We can express the acceleration as \(a = \frac{F}{m}\). Substitute the values: \(F = 5\, \mathrm{N}\) and \(m = 2\,\mathrm{kg}\).
05

Calculate the Acceleration

Substitute the force and mass into the equation for acceleration: \[ a = \frac{5\, \mathrm{N}}{2\, \mathrm{kg}} = 2.5\, \mathrm{m/s^2} \].
06

Select the Correct Answer

Compare the calculated acceleration \(2.5\, \mathrm{m/s^2}\) with the provided options. \(2.5 = \frac{5}{2} \), so the correct answer is option d.

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Key Concepts

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

Momentum
Momentum is a key concept in physics that represents the quantity of motion an object has. It is a vector quantity that depends on both the mass of the object and its velocity. The formula is given by
  • Momentum (\(p\)) = Mass (\(m\)) × Velocity (\(v\))
The principles of momentum are crucial in cases like our exercise with the water jet and the metal block. When the water jet strikes the block, it transfers momentum to the block. As per Newton's Second Law, when a force is applied it causes a change in momentum.

Here, the force on the block equals the rate at which the water changes its momentum. The water jet has a momentum change per second, or force, which is determined by multiplying the mass flow rate by the velocity of the water. This gives us a force of \(5 \, \text{N}\), as calculated in the solution. Understanding momentum helps us understand how forces cause things to move and accelerate.
Frictionless Plane
A frictionless plane is a theoretical concept used in physics problems to simplify the effects of friction. In real world scenarios, some surface interaction is always present, but assuming a frictionless surface means that we don't have to account for any frictional forces opposing the movement.

This is beneficial because it allows you to focus solely on other forces at play, like the force of the water jet striking the block in our exercise. Without the resistance of friction, any external non-perpendicular force applied causes motion. This perfectly highlights Newton's First Law, where an object at rest stays at rest and an object in motion stays in motion unless acted upon by an external force. Here, the force from the water applies directly and entirely to move the block, simplifying calculations.
Acceleration
Acceleration is the change in velocity of an object over time. It's a crucial concept in understanding how objects move and is defined by the formula
  • \(a = \frac{F}{m}\)
where \(a\) is acceleration, \(F\) is force, and \(m\) is mass. In our problem, the jet's force on the metal block causes the block to accelerate over the frictionless plane.

According to Newton's Second Law, acceleration of an object depends on two variables: the net force acting upon the object and the mass of the object. Higher force or lower mass leads to greater acceleration. In the problem, using the calculated force of \(5 \, \text{N}\) and the mass of the block at \(2 \, \text{kg}\), the block's acceleration is calculated to be \(2.5 \, \text{m/s}^2\). This shows the direct relationship between force, mass, and acceleration and illustrates how external forces, like our water jet, cause objects to move.

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Most popular questions from this chapter

A balloon of mass \(M\) is descending at a constant acceleration \(\alpha\). When a mass \(m\) is released from the balloon it starts rising with the same acceleration \(a\). Assuming that its volume does not change, what is the value of \(m\) ? a. \(\frac{\alpha}{\alpha+g} M\) b. \(\frac{2 \alpha}{\alpha+g} M\) c. \(\frac{\alpha+g}{\alpha} M\) d. \(\frac{\alpha+g}{2 \alpha} M\)

A block of mass \(m\) is lying on a wedge having inclination angle \(\alpha=\tan ^{-1}\left(\frac{1}{5}\right)\). Wedge is moving with a constant acceleration \(a=2 \mathrm{~m} / \mathrm{s}^{2} .\) The minimum value of coefficient of friction \(\mu\), so that \(m\) remains stationary wr.t. to wedge is a. \(2 / 9\) b. \(5 / 12\) c. \(1 / 5\) d. \(2 / 5\)

A box of mass \(8 \mathrm{~kg}\) is placed on a rough inclined plane of inclination \(\theta\). Its downward motion can be prevented by applying an upward pull \(F\) and it can be made to slide upwards by applying a force \(2 F\). The coefficient of friction between the box and the inclined plane is a. \((\tan \theta) / 3\) 1\. \(3 \tan \theta\) c. \((\tan \theta) / 2\) d. \(2 \tan \theta\)

A body of mass \(m\) is launched up on a rough inclined plane making an angle \(45^{\circ}\) with horizontal. If the time of ascent is half of the time of descent, the frictional coefficient between plane and body is a. \(\frac{2}{5}\) b. \(\frac{3}{5}\) c. \(\frac{3}{4}\) d. \(\frac{4}{5}\)

The upper half of an inclined plane with inclination \(\phi\) is perfectly smooth while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom if the coefficient of friction for the lower half is given by a. \(2 \tan \phi\) b. \(\tan \phi\) c. \(2 \sin \phi\) d. \(2 \cos \phi\)

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