/*! 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 73 A wooden block with mass \(1.50 ... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 73

A wooden block with mass \(1.50 \mathrm{~kg}\) is placed against a compressed spring at the bottom of an incline of slope \(30.0^{\circ}\) (point \(A\) ). When the spring is released, it projects the block up the incline. At point \(B,\) a distance of \(6.00 \mathrm{~m}\) up the incline from \(A\), the block is moving up the incline at \(7.00 \mathrm{~m} / \mathrm{s}\) and is no longer in contact with the spring. The coefficient of kinetic friction between the block and the incline is \(\mu_{k}=0.50\) The mass of the spring is negligible. Calculate the amount of potential energy that was initially stored in the spring.

Short Answer

Expert verified
The amount of potential energy that was initially stored in the spring is approximately \(119.205 \mathrm{J}\).

Step by step solution

01

Calculate the Potential Energy at B

The potential energy at point B can be calculated by using the formula for potential energy. Here, the mass of the block is \(1.50 \mathrm{~kg}\), acceleration due to gravity is \(9.81 \mathrm{~m/s^2}\), and height at point B is \(6.00 \mathrm{~m} * sin(30^{\circ})\). Using the formula \(PE = m*g*h\), the potential energy turns out to be \(PE = 1.50 \mathrm{~kg} * 9.81 \mathrm{~m/s^2} * 6.00 \mathrm{~m} * sin(30^{\circ}) = 44.145 \mathrm{J}\).
02

Calculate the Kinetic Energy at B

The kinetic energy at point B can be calculated using the formula for kinetic energy. Here the mass of the block is \(1.50 \mathrm{kg}\), and it's given that the velocity at point B is \(7.00 \mathrm{m/s}\). Using the formula \(KE = \frac{1}{2} * m * v^2\), the kinetic energy turns out to be \(KE = 0.5 * 1.50 \mathrm{~kg} * (7.00 \mathrm{~m/s})^2 = 36.75 \mathrm{~J}\).
03

Calculate the Work done by Friction

The work done by friction can be calculated using the formula \(W = f*d\). Here the force of friction can be calculated using the formula \(f = \mu_{k} * m * g * cos(\theta)\), where \(\mu_{k}\) is the coefficient of kinetic friction which is given as \(0.50\), \(m\) is the mass of the block which is \(1.50 \mathrm{~kg}\), \(g\) is the acceleration due to gravity \(9.81 \mathrm{~m/s^2}\), and the cosine of the angle of incline \(cos(30^{\circ})\). The distance the block moved d is given as \(6.00 \mathrm{~m}\). So, the work done by friction turns out to be \(W = \mu_{k} * m * g * cos(\theta) * d = 0.50 * 1.50 \mathrm{~kg} * 9.81 \mathrm{~m/s^2} * cos(30^{\circ}) * 6.00 \mathrm{~m} = 38.31 \mathrm{~J}\)
04

Calculate the Initial Potential Energy in the Spring

The total energy at B is the sum of the potential energy, the kinetic energy and the work done by friction (which is subtracted because it represents energy loss). This total must also be the amount of potential energy that was initially stored in the spring, since energy must be conserved. So, using the formula \(PE_{initial} = KE + PE + W\), it turns out to be \(PE_{initial} = 36.75 \mathrm{~J} + 44.145 \mathrm{~J} + 38.31 \mathrm{~J} = 119.205 \mathrm{~J}\)

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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 an object possesses due to its motion. It's an essential concept in physics and can be calculated using the formula:
  • \( KE = \frac{1}{2} m v^2 \)
Here, \(m\) represents the mass of the object, and \(v\) is its velocity. When analyzing the motion of the wooden block on the incline, we find that at point B, it is moving at \(7.00\, \mathrm{m/s}\). With the mass of \(1.50\, \mathrm{kg}\), the kinetic energy is calculated to be \(36.75\, \mathrm{J}\). This energy shows how fast the block is moving up the incline and plays a crucial role in determining the total mechanical energy of the system.
Potential Energy
Potential energy is the stored energy in an object due to its position within a force field, like gravity. For the wooden block on the incline, gravitational potential energy increases as the block moves up the slope.
  • The formula used is: \( PE = m \cdot g \cdot h \)
Where \(m\) is mass, \(g\) is the acceleration due to gravity, and \(h\) is the height. At point B, the block rises to a height calculated from the incline length and angle, giving us a potential energy of \(44.145\, \mathrm{J}\). This energy reflects the work done against gravity to elevate the block.
Work Done by Friction
Work done by friction is the energy lost as the block moves up the incline due to frictional forces between the block and the surface. The force of friction depends on the normal force and the coefficient of kinetic friction.
  • The formula to calculate frictional work is: \(W = f \cdot d\)
  • Where \(f = \mu_{k} \cdot m \cdot g \cdot \cos(\theta)\)
Here, \(\mu_{k}\) is the friction coefficient, and \(d\) is the distance moved. With a coefficient of \(0.50\) and the given data, the work done by friction is \(38.31\, \mathrm{J}\). This energy is subtracted from total energy calculations as it represents energy dissipated due to friction.
Spring Mechanics
Spring mechanics involve potential energy stored in a spring when it is compressed or stretched. This stored energy is released when the spring returns to its original shape. In the exercise, the spring's energy is initially what propels the block up the incline.
  • The energy stored in the spring explains the initial motion and must be equal to the total energy at point B.
  • According to the conservation of energy: \[PE_{initial} = KE + PE + W\]
Adding the kinetic, potential, and frictional work gives us the spring's initial potential energy of \(119.205\, \mathrm{J}\). This is the energy stored when the spring was compressed, showing how mechanical energy is conserved and transferred within the system.

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Most popular questions from this chapter

The Great Sandini is a \(60 \mathrm{~kg}\) circus performer who is shot from a cannon (actually a spring gun). You don't find many men of his caliber, so you help him design a new gun. This new gun has a very large spring with a very small mass and a force constant of \(1100 \mathrm{~N} / \mathrm{m}\) that he will compress with a force of \(4400 \mathrm{~N}\). The inside of the gun barrel is coated with Teflon, so the average friction force will be only \(40 \mathrm{~N}\) during the \(4.0 \mathrm{~m}\) he moves in the barrel. At what speed will he emerge from the end of the barrel, \(2.5 \mathrm{~m}\) above his initial rest position?

\(\mathrm{CP}\) A \(0.300 \mathrm{~kg}\) potato is tied to a string with length \(2.50 \mathrm{~m}\) and the other end of the string is tied to a rigid support. The potato is held straight out horizontally from the point of support, with the string pulled taut, and is then released. (a) What is the speed of the potato at the lowest point of its motion? (b) What is the tension in the string at this point?

Two blocks are attached to either end of a light rope that passes over a light, frictionless pulley suspended from the ceiling. One block has mass \(8.00 \mathrm{~kg},\) and the other has mass \(6.00 \mathrm{~kg}\). The blocks are released from rest. (a) For a \(0.200 \mathrm{~m}\) downward displacement of the \(8.00 \mathrm{~kg}\) block, what is the change in the gravitational potential energy associated with each block? (b) If the tension in the rope is \(T\), how much work is done on each block by the rope? (c) Apply conservation of energy to the system that includes both blocks. During the \(0.200 \mathrm{~m}\) downward displacement, what is the total work done on the system by the tension in the rope? What is the change in gravitational potcntial energy associated with the system? Use energy conservation to find the speed of the \(8.00 \mathrm{~kg}\) block after it has descended \(0.200 \mathrm{~m} .\)

The food calorie, equal to \(4186 \mathrm{~J},\) is a measure of how much energy is released when the body metabolizes food. A certain fruit-and-cereal bar contains 140 food calories. (a) If a 65 kg hiker eats one bar, how high a mountain must he climb to "work off" the calories, assuming that all the food energy goes into increasing gravitational potential energy? (b) If, as is typical, only \(20 \%\) of the food calories go into mechanical energy, what would be the answer to part (a)? (Note: In this and all other problems, we are assuming that \(100 \%\) of the food calories that are eaten are absorbed and used by the body. This is not true. A person's "metabolic efficiency" is the percentage of calories eaten that are actually used; the body eliminates the rest. Metabolic efficiency varies considerably from person to person.

BIO Tendons. Tendons are strong elastic fibers that attach muscles to bones. To a reasonable approximation, they obey Hooke's law. In laboratory tests on a particular tendon, it was found that, when a \(250 \mathrm{~g}\) object was hung from it, the tendon stretched \(1.23 \mathrm{~cm}\). (a) Find the force constant of this tendon in \(\mathrm{N} / \mathrm{m}\). (b) Because of its thickness, the maximum tension this tendon can support without rupturing is 138 N. By how much can the tendon stretch without rupturing. and how much energy is stored in it at that point?
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