/*! 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 72 The block of mass \(m\) initiall... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 72

The block of mass \(m\) initially at \(x=0\) is acted upon by a horizontal force \(F=a-b x^{2}\) (where \(a>\mu m g\) ), as shown in the figure. The co-efficient of friction between the surfaces of contact is \(\mu\). The net work done on the block is zero, if the block travels a distance of (1) \(x=\sqrt{\frac{3(a-\mu m g)}{b}}\) (2) \(x=\sqrt{\frac{3(a+\mu m g)}{b}}\) (3) \(x=\sqrt{\frac{2(a-\mu m g)}{b}}\) (4) \(x=\sqrt{\frac{2(a+\mu m g)}{b}}\)

Short Answer

Expert verified
The block travels a distance of \(x = \sqrt{\frac{3(a - \mu mg)}{b}}\), which is option (1).

Step by step solution

01

Analyze the Problem

Given a block of mass \(m\) on a surface with a coefficient of friction \(\mu\) and acted upon by a force \(F = a - bx^2\), we need to find the distance \(x\) for which the net work done on the block is zero.
02

Determine Forces Involved

The applied force is \(F = a - bx^2\), and the frictional force is \(f = \mu mg\). The block will move until the work done by the net force (applied force minus friction force) is zero.
03

Set Up the Energy Balance

The work done by the net force \(F - f\) over the distance \(x\) must be zero. Hence, we can write the equation as \(\int_0^x (a - bx^2 - \mu mg) \, dx = 0\).
04

Solve the Integral

The integral \(\int_0^x (a - bx^2 - \mu mg) \, dx\) simplifies to \([ax - \frac{b}{3}x^3 - \mu mg \, x]_0^x = 0\). Substituting the limits, we obtain \(ax - \frac{b}{3}x^3 - \mu mg \, x = 0\).
05

Solve for \(x\)

Rearrange the equation: \(ax - \mu mg \, x = \frac{b}{3}x^3\). Dividing by \(x\) (since \(x eq 0\)), we have \(a - \mu mg = \frac{b}{3}x^2\) \(\Rightarrow x^2 = \frac{3(a - \mu mg)}{b}\).
06

Final Answer Computation

By solving the equation for \(x\), we get \(x = \sqrt{\frac{3(a - \mu mg)}{b}}\). This corresponds to option (1).

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Key Concepts

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

Frictional Force
Frictional force is an important concept in physics that often comes into play when discussing the movement of objects. It is the force that resists the motion of an object relative to a surface. This force acts in the opposite direction of the applied force that is attempting to move the object. In our exercise scenario, the block of mass \(m\) is experiencing a frictional force due to the contact between its surface and the surface it's on. The magnitude of this frictional force is determined by the coefficient of friction \(\mu\) and the normal force, which, in this case, is the gravitational force \(mg\). Hence, the frictional force \(f\) can be calculated as:
  • \(f = \mu \cdot mg\)
The frictional force plays a crucial role in determining how far the block can travel before it comes to a stop. Since the net work done is the difference between the work done by the applied force and the work done against friction, understanding this relationship is key to solving such problems.
Variable Force
In contrast to constant forces, a variable force changes as the position of the object changes. In this exercise, the horizontal force acting on the block is not constant. It has been defined as \(F = a - bx^2\), which implies that the force varies with the position \(x\). It鈥檚 essential to understand how variable forces affect motion differently from constant forces.
Because the force is dependent on position, as the block moves along the surface, the force exerted on it decreases quadratically with distance. This means that as \(x\) increases, the value of \(bx^2\) becomes more significant, reducing the overall force \(F\). Therefore, the opposition due to friction eventually causes the net force to drop to zero. Understanding the behavior of variable forces is crucial in determining the block鈥檚 motion and the distance at which the net work done becomes zero.
Integral Calculus
Integral calculus is a branch of mathematics that deals with finding quantities from their rates of change. It is used to calculate accumulated values such as areas, volumes, or, in this case, work done over a distance. In the given exercise, to calculate the net work done on the block, we need to evaluate the integral of the net force over the distance \(x\).
The work-energy principle helps us set up the integral as:
  • \[ \int_0^x (a - bx^2 - \mu mg) \, dx = 0 \]
This calculation helps us find the specific point \(x\) where the work done by the net force is zero. By computing the integral, we accumulate the effects of the variable force and the frictional force over the block's journey. Solving the integral gives us the formula to determine the distance \(x\) the block travels before stopping due to the balance of forces.
Understanding integral calculus is key to solving complex physics problems where forces are not constant and vary over distances.

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Most popular questions from this chapter

A single conservative force \(F(x)\) acts on a \(1.0-\mathrm{kg}\) particle that moves along the \(x\)-axis. The potential energy \(U(x)\) is given by \(U(x)=20+(x-2)^{2}\) where \(x\) is in meters. At \(x=5.0 \mathrm{~m}\), the particle has a kinetic energy of \(20 \mathrm{~J}\). The maximum kinetic energy of the particle and the value of \(x\) at which maximum kinetic energy occurs are (1) \(29 \mathrm{~J}, 0 \mathrm{~m}\) (2) \(49 \mathrm{~J}, 0 \mathrm{~m}\) (3) \(49 \mathrm{~J}, 2 \mathrm{~m}\) (4) \(29 \mathrm{~J}, 2 \mathrm{~m}\)

One end of a spring of force constant \(k_{1}\) is attached to the ceiling of an elevator. A block of mass \(1.5 \mathrm{~kg}\) is attached to the other end. Another spring of force constant \(k_{2}\) is attached to the bottom of the mass and to the floor of the elevator as shown in the figure. At equilibrium, the deformation in both the springs is equal and is \(40 \mathrm{~cm}\). If the elevator moves with constant acceleration upward, the additional deformation in both the springs is \(8 \mathrm{~cm}\). Find the elevator's acceleration \(\left(g=10 \mathrm{~m} \mathrm{~s}^{-2}\right)\).

A ring of mass \(m=1 \mathrm{~kg}\) can slide over a smooth vertical rod. A light string attached to the ring passing over a smooth fixed pulley at a distance of \(L=0.7 \mathrm{~m}\) from the rod is shown in the figure. At the other end of the string, mass \(M=5 \mathrm{~kg}\) is attached, lying over a smooth fixed inclined plane of inclination angle \(37^{\circ}\). The ring is held in level with the pulley and released. Determine the velocity of ring (in \(\mathrm{m} / \mathrm{s}\) ) when the string mat : an angle \(\left(\alpha=37^{\circ}\right)\) with the horizontal. \(\left[\sin 37^{\circ}=0.6\right]\)

A bus can be stopped by applying a retarding force \(F\) when it is moving with speed \(v\) on a level road. The distance covered by it before coming to rest is \(s\). If the load of the bus increases by \(50 \%\) because of passengers, for the same speed and same retarding force, the distance covered by the bus to come to rest shall be (1) \(1.55\) (2) \(2 s\) (3) \(1 s\) (4) \(2.5 \mathrm{~s}\)

A \(1.5-\mathrm{kg}\) block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of \(x\)-axis is applied to the block. The force is given by \(\vec{F}=\left(4-x^{2}\right) \vec{i} \mathrm{~N}\), where \(x\) is in meter and the initial position of the block is \(x=0\). The maximum positive displacement \(x\) is (1) \(2 \sqrt{3} \mathrm{~m}\) (2) \(2 \mathrm{~m}\) (3) \(4 \mathrm{~m}\) (4) \(\sqrt{2} \mathrm{~m}\)
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