A 4.0 -kg block is released from rest at the top of a...

A 4.0 -kg block is released from rest at the top of a frictionless plane of length $8.0 \mathrm{m}$ that is inclined at an angle of $15^{\circ}$ to the horizontal. A cord is attached to the block and trails along behind it. When the block reaches a point $5.0 \mathrm{m}$ along the incline from the top, someone grasps the cord and exerts a constant tension parallel to the incline. The tension is such that the block just comes to rest when it reaches the bottom of the incline. (The person's force is a nonconservative force.) What is this constant tension? Solve the problem twice, once using work and energy and again using Newton's laws and the equations for constant acceleration. Which method do you prefer? A 4.0 -kg block is released from rest at the top of a frictionless plane of length $8.0 \mathrm{m}$ that is inclined at an angle of $15^{\circ}$ to the horizontal. A cord is attached to the block and trails along behind it. When the block reaches a point $5.0 \mathrm{m}$ along the incline from the top, someone grasps the cord and exerts a constant tension parallel to the incline. The tension is such that the block just comes to rest when it reaches the bottom of the incline. (The person's force is a nonconservative force.) What is this constant tension? Solve the problem twice, once using work and energy and again using Newton's laws and the equations for constant acceleration. Which method do you prefer?

A 4.0 -kg block is released from rest at the top of a frictionless plane of length $8.0 \mathrm{m}$ that is inclined at an angle of $15^{\circ}$ to the horizontal. A cord is attached to the block and trails along behind it. When the block reaches a point $5.0 \mathrm{m}$ along the incline from the top, someone grasps the cord and exerts a constant tension parallel to the incline. The tension is such that the block just comes to rest when it reaches the bottom of the incline. (The person's force is a nonconservative force.) What is this constant tension? Solve the problem twice, once using work and energy and again using Newton's laws and the equations for constant acceleration. Which method do you prefer?
A 4.0 -kg block is released from rest at the top of a frictionless plane of length $8.0 \mathrm{m}$ that is inclined at an angle of $15^{\circ}$ to the horizontal. A cord is attached to the block and trails along behind it. When the block reaches a point $5.0 \mathrm{m}$ along the incline from the top, someone grasps the cord and exerts a constant tension parallel to the incline. The tension is such that the block just comes to rest when it reaches the bottom of the incline. (The person's force is a nonconservative force.) What is this constant tension? Solve the problem twice, once using work and energy and again using Newton's laws and the equations for constant acceleration. Which method do you prefer?

Key Concepts

Decomposition of Forces on an Inclined Plane

In problems involving inclined planes, gravitational forces are often decomposed into components parallel and perpendicular to the plane. This decomposition is crucial for accurately applying both energy and Newton's law methods, as the parallel component determines the motion along the incline while the perpendicular component is balanced by the normal force.

Kinematics of Constant Acceleration

Under conditions of constant acceleration, kinematic equations provide relationships among displacement, velocity, time, and acceleration. These equations are useful in problems where forces yield uniform acceleration, allowing for the determination of unknown forces or accelerations by relating them to measured displacements and velocities.

Work-Energy Theorem

This concept relates the work done by all forces (including nonconservative forces) to the change in kinetic energy of an object. It provides an energy-based approach to problem solving, where changes in potential and kinetic energy are balanced by the work done by forces like tension, allowing for solutions without directly calculating acceleration at each step.

Newton's Second Law

Newton's Second Law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. This principle underlies the direct force analysis method, where free-body diagrams are used to resolve forces along the incline, and kinematic equations are applied to relate force, mass, acceleration, and displacement.

Nonconservative Forces

Nonconservative forces, such as tension or friction, dissipate energy or add energy to a system in a way that depends on the path taken. When using energy methods, the work done by these forces must be carefully accounted for because they change the total mechanical energy of the system, unlike conservative forces which only convert energy between potential and kinetic forms.

Transcript

00:01 In this exercise, we have a block that has a mass of 4 kilograms that is on the top of an inclined plane that has a length of 8 meters and makes an angle of 15 degrees with horizontal.

00:15 The block is initially at rest and then it starts to slide down the plane, and the block has a cord that trails along behind it.

00:25 And then, after the block slides down a distance d of 5 meters from the top of the inclined plane, someone takes a cord that's behind the block and starts to exert a certain force over the cord, such that the block stops exactly at the bottom of the inclined plane.

00:51 And our goal is to calculate the force that the person exerts over the cord in two different ways.

01:00 First, we have to use conservation of energy methods in order to find it, and then we have to use newton's laws and the concept of acceleration in order to find it.

01:12 So first of all, in order to find it using energy methods, i'm going to use the concept that the work that's done over a body, body is equal to the variation in its mechanical energy.

01:35 Now notice that initially the body is, when it's at the top of the inclined plane, it has a mechanical energy, initial mechanical energy of mgh, okay, where h is equal to 8 times sine of 15 degrees.

02:07 This is the initial mechanical energy and the block keeps this energy for 5 meters until someone takes the cord and starts to exert a certain force over the cord.

02:24 Then the final energy of the block is zero because at the bottom of the inclined plane h is equal to zero and the block has no kinetic energy.

02:39 So the work is equal to the final, actually the should be ef, is equal to the variation in energy, so it's the final energy minus the initial energy.

02:51 So this is minus mgh.

02:56 And the work is the force f times the distance l along which the force is exerted, and this is equal to minus mgh.

03:12 So the force is minus mgah divided by l.

03:21 M is 4 kilograms.

03:24 G is 9 meters per second squared.

03:29 H is 8, sine of 15, and l is 3 meters.

03:39 So f is equal to 27 newton.

03:49 Then we can solve it using force method, a force method.

04:03 And basically, notice that we have to draw the force diagram on the block.

04:12 So here we have the block.

04:15 We have the gravitational force that acts upon the block, but we can divide it into two components, one that is tangential to the inclined plane, and it's equal to mg sine theta.

04:27 X .0 .0 .0 .0 .0 .0 .0.

04:30 The other one, that is m .g...

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