/*! 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 82 A \(4.00 \mathrm{~kg}\) block is... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(4.00 \mathrm{~kg}\) block is pulled up a frictionless inclined plane by a \(50.0 \mathrm{~N}\) force that is parallel to the plane, starting from rest. The normal force on the block from the plane has magnitude \(13.41 \mathrm{~N}\). What is the block's speed when its displacement up the ramp is \(3.00 \mathrm{~m} ?\)

Short Answer

Expert verified
The block's speed is approximately 8.66 m/s after moving 3.00 m up the ramp.

Step by step solution

01

Identify Known Values

We know the following values:- Mass of the block, \( m = 4.00 \mathrm{~kg} \)- Force applied parallel to the plane, \( F = 50.0 \mathrm{~N} \)- Normal force, \( N = 13.41 \mathrm{~N} \)- Displacement along the plane, \( s = 3.00 \mathrm{~m} \)- Initial velocity of the block, \( u = 0 \mathrm{~m/s} \)We need to find the block's speed after moving a displacement \( s = 3.00 \mathrm{~m} \).
02

Calculate the Work Done on the Block

The work done \( W \) by the force \( F \) is found using the formula:\[ W = F \times s \]Substituting the known values:\[ W = 50.0 \mathrm{~N} \times 3.00 \mathrm{~m} = 150.0 \mathrm{~J} \]
03

Apply Work-Energy Principle

According to the work-energy principle, the work done on the block equals its change in kinetic energy:\[ \Delta KE = W \]The change in kinetic energy \( \Delta KE \) is:\[ \Delta KE = \frac{1}{2}mv^2 - \frac{1}{2}mu^2 \]Since it starts from rest, \( u = 0 \), so \( \Delta KE = \frac{1}{2}mv^2 \).Thus:\[ \frac{1}{2}mv^2 = 150.0 \mathrm{~J} \]
04

Solve for Final Speed

Substitute \( m = 4.00 \mathrm{~kg} \) into the equation and solve for \( v \):\[ \frac{1}{2} \times 4.00 \mathrm{~kg} \times v^2 = 150.0 \mathrm{~J} \]\[ 2.00 \mathrm{~kg} \times v^2 = 150.0 \mathrm{~J} \]\[ v^2 = \frac{150.0 \mathrm{~J}}{2.00 \mathrm{~kg}} = 75.0 \mathrm{~m^2/s^2} \]\[ v = \sqrt{75.0 \mathrm{~m^2/s^2}} \]\[ v = 8.66 \mathrm{~m/s} \]
05

Conclusion

The speed of the block after being displaced \(3.00 \mathrm{~m}\) up the ramp is approximately \(8.66 \mathrm{~m/s}\).

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Key Concepts

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

Kinetic Energy
Kinetic energy (\( KE \)) is the energy that an object possesses due to its motion. It is a crucial concept to understand when studying physics, as it helps us comprehend how motion translates into energy. The kinetic energy of an object depends on two factors:
  • The mass of the object (\( m \))
  • The speed of the object (\( v \))
The formula to calculate kinetic energy is:\[ KE = \frac{1}{2}mv^2 \]This means that the energy increases rapidly with speed because it is proportional to the square of the velocity. To move an object, we need to do work on it, and this work converts to kinetic energy as the object starts to move. For example, in our exercise, the block initially at rest gains kinetic energy as it is pulled up the incline, increasing its speed as work is done on it by the applied force.
Work Done on an Inclined Plane
The concept of work describes the energy transferred to an object via a force applied over a distance. For an inclined plane, calculating work done is a bit unique but follows the same principle. Work (\( W \)) is calculated with the formula:\[ W = F \times s \]where \( F \) is the force applied parallel to the slope, and \( s \) is the displacement along the incline. For tasks involving inclined planes, the force applied is often aligned with the slope, making calculations straightforward. In the described exercise, a force of 50 N is applied, moving the block 3 meters up the incline, resulting in 150 J of work done on the block. This work is pivotal in increasing the block's kinetic energy and thus its speed.
Newton's Laws of Motion
Newton's Laws of Motion are foundational principles that govern how objects move and interact. Here's a brief overview:
  • First Law: An object at rest stays at rest, and an object in motion stays in motion unless acted upon by a net external force. This law highlights the importance of forces in changing an object's state.
  • Second Law: The acceleration of an object depends on the net force acting on it and its mass (\( F = ma \)). This principle is vital in understanding how different forces affect acceleration, explaining why the block accelerates up the plane.
  • Third Law: For every action, there is an equal and opposite reaction. This law explains the interactions between the block and the inclined plane surface, such as the normal force balancing gravitational forces perpendicular to the slope.
In our given exercise, Newton's Second Law plays a crucial role where the 50 N force applied on the block leads to its acceleration and motion up the plane. This interaction illustrates how forces govern movement and how energy principles complement these laws to describe motion comprehensively.

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Most popular questions from this chapter

A force of \(5.0 \mathrm{~N}\) acts on a \(15 \mathrm{~kg}\) body initially at rest. Compute the work done by the force in (a) the first, (b) the second, and (c) the third seconds and (d) the instantaneous power due to the force at the end of the third second.

A luge and its rider, with a total mass of \(85 \mathrm{~kg}\), emerge from a downhill track onto a horizontal straight track with an initial speed of \(37 \mathrm{~m} / \mathrm{s} .\) If a force slows them to a stop at a constant rate of \(2.0 \mathrm{~m} / \mathrm{s}^{2}\) (a) what magnitude \(F\) is required for the force, (b) what distance \(d\) do they travel while slowing, and (c) what work \(W\) is done on them by the force? What are (d) \(F,(\mathrm{e}) d,\) and \((\mathrm{f}) \mathrm{W}\) if they, instead, slow at \(4.0 \mathrm{~m} / \mathrm{s}^{2} ?\)

The loaded cab of an elevator has a mass of \(3.0 \times 10^{3} \mathrm{~kg}\) and moves \(210 \mathrm{~m}\) up the shaft in \(23 \mathrm{~s}\) at constant speed. At what average rate does the force from the cable do work on the cab?

A fully loaded, slow-moving freight elevator has a cab with a total mass of \(1200 \mathrm{~kg}\), which is required to travel upward \(54 \mathrm{~m}\) in \(3.0 \mathrm{~min},\) starting and ending at rest. The elevator's counterweight has a mass of only \(950 \mathrm{~kg}\), and so the elevator motor must help. What average power is required of the force the motor exerts on the cab via the cable?

A \(250 \mathrm{~g}\) block is dropped onto a relaxed vertical spring that has a spring constant of \(k=\) \(2.5 \mathrm{~N} / \mathrm{cm}\) (Fig. \(7-46)\). The block becomes attached to the spring and compresses the spring \(12 \mathrm{~cm}\) before momentarily stopping. While the spring is being compressed, what work is done on the block by (a) the gravitational force on it and (b) the spring force? (c) What is the speed of the block just before it hits the spring? (Assume that friction is negligible.) (d) If the speed at impact is doubled, what is the maximum compression of the spring?

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