/*! 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 37 A \(40 \mathrm{~kg}\) slab rests... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 37

A \(40 \mathrm{~kg}\) slab rests on a frictionless floor. A \(10 \mathrm{~kg}\) block rests on top of the slab (Fig. 6-58). The coefficient of static friction \(\mu^{\text {stat }}\) between the block and the slab is \(0.60\), whereas their kinetic friction coefficient \(\mu^{\text {kin }}\) is \(0.40\). The \(10 \mathrm{~kg}\) block is pulled by a horizontal force with a magnitude of \(100 \mathrm{~N}\). What are the resulting accelerations of (a) the block and (b) the slab?

Short Answer

Expert verified
The block accelerates at 6.08 m/s^2 and the slab accelerates at 0.98 m/s^2.

Step by step solution

01

Determine Maximum Static Friction

Calculate the maximum static friction force using the formula:\[ f_{\text{max}} = \text{static friction coefficient} \times \text{normal force} \] The normal force is given by the weight of the block, which is \[ f_{\text{max}} = \text{0.60} \times (10 \text{~kg} \times 9.8 \text{~m/s}^2) = 58.8 \text{~N} \]
02

Compare Applied Force with Maximum Static Friction

Compare the applied force to the maximum static friction to determine whether the block will move relative to the slab. The applied force is 100 N, which is greater than the maximum static friction force of 58.8 N. Thus, the block will slide over the slab.
03

Calculate Kinetic Friction Force

Since the block will slide, calculate the kinetic friction force using the kinetic friction coefficient:\[ f_{\text{kin}} = \text{kinetic friction coefficient} \times \text{normal force} \] The normal force remains the same, so \[ f_{\text{kin}} = 0.40 \times (10 \text{~kg} \times 9.8 \text{~m/s}^2) = 39.2 \text{~N}\]
04

Determine Acceleration of Block

Apply Newton's second law to the block. The net force acting on the block is the difference between the applied force and the kinetic friction force:\[ F_{\text{net}} = 100 \text{~N} - 39.2 \text{~N} = 60.8 \text{~N}\] Use \( F = ma \) to find the acceleration of the block:\[ a_{\text{block}} = \frac{F_{\text{block, net}}}{m_{\text{block}}} = \frac{60.8 \text{~N}}{10 \text{~kg}} = 6.08 \text{~m/s}^2\]
05

Determine Acceleration of Slab

The only horizontal force acting on the slab is the kinetic friction force from the block. Use Newton's second law to determine the acceleration of the slab:\[ a_{\text{slab}} = \frac{F_{\text{kin}}}{m_{\text{slab}}} = \frac{39.2 \text{~N}}{40 \text{~kg}} = 0.98 \text{~m/s}^2 \]

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Key Concepts

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

Newton's Laws of Motion
Newton's Laws of Motion are fundamental principles in physics that describe the relationship between the motion of an object and the forces acting on it. In problems like these, Newton's second law is especially useful. It states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Mathematically, it's expressed as:\[ F = ma \] where \(F\) is the net force, \(m\) is the mass, and \(a\) is the acceleration.
In this exercise, we apply Newton's second law to both the block and the slab to find their accelerations.
Frictional Forces
Frictional forces resist the relative motion between two surfaces in contact. There are two main types: static friction and kinetic friction. Static friction prevents objects from starting to move, whereas kinetic friction opposes the motion of objects that are already moving. In this problem, friction acts between the block and the slab and is crucial in determining whether the block will slide or stay put on the slab.
Kinetic Friction
Kinetic friction comes into play when objects are sliding against each other. The kinetic frictional force is calculated using the coefficient of kinetic friction (\( \mu_{\text{kin}} \)) and the normal force. It's given by the equation: \[ f_{\text{kin}} = \mu_{\text{kin}} \times N \] Here, the normal force \(N\) is the weight of the block, which is the mass times the gravitational acceleration (\(9.8 \(m/s^2\)\)). In this exercise, the kinetic frictional force between the block and the slab slows down the block as it moves.
Static Friction
Static friction keeps objects at rest from moving. It is usually stronger than kinetic friction. The maximum static friction force can be calculated using the coefficient of static friction (\( \mu_{\text{stat}} \)) and the normal force as follows: \[ f_{\text{max}} = \mu_{\text{stat}} \times N \] In the given exercise, if the horizontal force applied to the block is less than this maximum static friction force, the block will not move relative to the slab. This threshold helps determine if the block will slide or stay stationary.
Normal Force
The normal force is the perpendicular force exerted by a surface on an object resting on it. For an object lying on a horizontal surface without any vertical acceleration, the normal force equals the object's weight. It can be calculated using the formula: \[ N = mg \] where \(m\) is the mass and \(g\) is the gravitational acceleration. In this exercise, knowing the normal force is essential to calculating both the static and kinetic frictional forces between the block and the slab.

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Most popular questions from this chapter

A young man is pushing a baby carriage at a constant velocity along a level street. A friend comes by to chat and the young man lets go of the carriage. It rolls on for a bit, slows, and comes to a stop. At time \(t=0\) the young man is walking with a constant velocity. At time \(t_{1}\) he releases the carriage. At time \(t_{2}\) the carriage comes to rest. Sketch qualitatively accurate (i.e., we don't care about the values but we do care about the shape) graphs of each of the following variables versus time: (a) position of the carriage, (b) velocity of the carriage, (c) acceleration of the carriage, (d) net force on the carriage, (e) force the man exerts on the carriage, (f) force of friction on the carriage. Be sure to note the important times \(t=0, t_{1}\), and \(t_{2}\) on the time axes of your graphs. Take the positive direction to be the direction in which the man was initially walking.

Suppose the coefficient of static friction between the road and the tires on a Formula One car is \(0.6\) during a Grand Prix auto race. What speed will put the car on the verge of sliding as it rounds a level curve of \(30.5 \mathrm{~m}\) radius?

A spaceship lifts off vertically from the Moon, where the freefall acceleration is \(1.6 \mathrm{~m} / \mathrm{s}^{2} .\) If the spaceship has an upward acceleration of \(1.0 \mathrm{~m} / \mathrm{s}^{2}\) as it lifts off, what is the magnitude of the force of the spaceship on its pilot, who weighs \(735 \mathrm{~N}\) on Earth?

George left the lights on in his truck while at a truck stop in Kansas and his battery went dead. Fortunately, his friend \(\mathrm{Al}\) is there, although \(\mathrm{Al}\) is driving his Geo Metro. Since the road is very flat, George is able to convince \(\mathrm{Al}\) to give his truck a long, slow push to get it up to 20 miles/hour. At this speed, George can engage the truck's clutch and the truck's engine should start up. (See Fig. 6-89.) (a) Al begins to push the truck. It takes him 5 minutes to get the truck up to a speed of 20 miles/hour. Draw separate free-body diagrams for the Geo and for the truck during the time that Al's Geo is pushing the truck. List all the horizontal forces in order by \(\mathrm{mag}\) nitude from largest to smallest. If any are equal, state that explicitly. Explain your reasoning. (b) If the truck is accelerating uniformly over the 5 minutes, how far does Al have to push the truck before George can engage the clutch? (c) Suppose the mass of the truck is \(4000 \mathrm{~kg}\), the mass of the car is \(800 \mathrm{~kg}\), and the coefficient of static friction between the vehicles and the road is \(0.1\). At one instant when they are trying to get the truck moving, the car is pushing the truck and exerting a force of \(1000 \mathrm{~N}\), but neither vehicle moves. What is the static frictional force between the truck and the road? Explain your reasoning.

A \(68 \mathrm{~kg}\) crate is dragged across a floor by pulling on a rope attached to the crate and inclined \(15^{\circ}\) above the horizontal. (a) If the coefficient of static friction is \(0.50\), what minimum force magnitude is required from the rope to start the crate moving? (b) If \(\mu^{\text {kin }}\) \(=0.35\), what is the magnitude of the initial acceleration of the crate?
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