/*! 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 49 A block of mass \(m\) is pulled ... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 49

A block of mass \(m\) is pulled by a constant power \(P\) placed on a rough horizontal plane. The friction co-efficient between the block and surface is \(\mu\). The maximum velocity of the block is (A) \(\frac{P}{m g}\) (B) \(\frac{P}{\mu m g}\) (C) \(\frac{\mu P}{m g}\) (D) Infinite

Short Answer

Expert verified
The maximum velocity of the block is given by the formula \(v = \frac{P}{\mu mg}\), which corresponds to answer choice (B).

Step by step solution

01

Determine the force due to the applied power

: Power is defined as the rate of work done with respect to time, and it is the product of force and velocity (Fv). In this case, the applied power is constant, which we denote as \(P\). We can write the equation for power as: \[ P = Fv \] Where \(F\) is the force applied and \(v\) is the velocity. We want to solve for the applied force, so we can rearrange this equation for \(F\): \[ F = \frac{P}{v} \]
02

Calculate the frictional force

: The frictional force acting on the block, which opposes the motion, can be calculated using the friction coefficient \(\mu\) and the normal force, which is equal to the weight of the block, i.e., \(mg\), where \(g\) is the acceleration due to gravity. The frictional force (\(F_\text{friction}\)) can be expressed as: \[ F_\text{friction} = \mu mg \]
03

Applying Newton's Second Law of motion

: Newton's Second Law states that the force acting on an object is equal to the mass of that object multiplied by its acceleration \(a\) (\(F = ma\)). At maximum velocity, there is no acceleration, so the net force acting on the block is zero. Therefore, the frictional force equals the applied force: \[ \frac{P}{v} = \mu mg \]
04

Solving for maximum velocity

: Now, we need to isolate \(v\) and find the maximum velocity. We rearrange the equation from step 3: \[ v = \frac{P}{\mu mg} \] Comparing this result with the answer choices, we find that our solution matches answer choice (B), which is: \[ \boxed{\text{(B) }\frac{P}{\mu mg}} \]

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Most popular questions from this chapter

A system consists of two identical cubes, each of mass \(3 \mathrm{~kg}\), linked together by a compressed weightless spring of force constant \(1000 \mathrm{~N} / \mathrm{m}\). The cubes are also connected by a thread which is burnt at a certain moment. At what minimum value of initial compression, \(x_{0}\) (in \(\mathrm{cm}\) ) of the spring will the lower cube bounce up after the thread is burnt through?

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A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a plane. It follows that (A) its velocity is constant. (B) its acceleration is constant. (B) its kinetic energy is constant. (D) it moves in a circular path.

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A block of mass \(m=0.1 \mathrm{~kg}\) is released from a height of \(4 \mathrm{~m}\) on a curved smooth surface. On the horizontal surface, path \(A B\) is smooth and path \(B C\) offers coefficient of friction \(\mu=0.1\). If the impact of block with the vertical wall at \(C\) be perfectly elastic, the total distance covered by the block on the horizontal surface before coming to rest will be: (take \(g=10 \mathrm{~m} / \mathrm{s}^{2}\) ) (A) \(29 \mathrm{~m}\) (B) \(49 \mathrm{~m}\) (C) \(59 \mathrm{~m}\) (D) \(109 \mathrm{~m}\)
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