# Tension of a Cord: Solving Homework Problem

• bewocc
In summary: Your Name]In summary, the problem involves a 4.0 kg block released from rest on a frictionless plane incline of length 8.0m and angle 15 degrees. A massless cord is attached to the block and someone grasps it at a point 5.0m down the incline, exerting a constant tension to bring the block to rest at the bottom. Using Newton's laws and constant acceleration kinematics, the tension is calculated to be -65N. To solve using work and energy, potential energy is calculated at two points and the work-energy theorem is used to find the tension, resulting in the same value of -65N.
bewocc

## Homework Statement

A 4.0 kg block is released from rest at the top of a frictionless plane of length 8.0m that is inclined at an angle of 15 degrees to the horizontal. A cord is attached to the block and trails along behind (assume the cord is massless). When the block reaches a point 5.0m 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. What is this constant tension? Solve the problem twice, once using work and energy and again using Newton's laws and constant acceleration kinematics.

## Homework Equations

(1) dfy = diy + viyt - 1/2gt^2
(2) dfx = dix + vixt +1/2at^2
(3) Fn - mgcos15 = 0
(4) mgsin15 - Ft = max

## The Attempt at a Solution

I'm really confused with this one. I've only attempted to solve the problem with Newton's laws and acceleration kinematics so far.
I used the first equation above to solve for time t = .5139s. Then I used the second equation to solve for ax = 18.796 m/s^2.
Lastly, I used the fourth equation to solve for Ft = -65N.
Can anyone confirm if this is correct? If it isn't, where did I go wrong?

Also, I don't know how to solve using work and energy. I know that work of a nonconservative force = change in kinetic energy + change in potential energy, but I don't know where to go from there.

Your solution using Newton's laws and constant acceleration kinematics is correct. You have correctly calculated the time and acceleration, and then used the equations to find the tension in the cord.

To solve using work and energy, you can use the following steps:

1. Calculate the potential energy at the top of the incline (PE1) and at the point where the block is grabbed (PE2). Use the equation PE = mgh, where m is the mass of the block, g is the acceleration due to gravity, and h is the height.

2. Calculate the change in potential energy (ΔPE = PE2 - PE1).

3. Use the work-energy theorem, which states that the work done by all forces on an object is equal to the change in kinetic energy (KE) of the object. In this case, the only force doing work is the tension in the cord, so we can write: W = FtΔx = ΔKE, where Ft is the tension, Δx is the distance traveled, and ΔKE is the change in kinetic energy.

4. Since the block starts from rest, the initial kinetic energy (KE1) is zero. Therefore, the final kinetic energy (KE2) is equal to the change in kinetic energy (ΔKE).

5. Use the equation for kinetic energy, KE = 1/2mv^2, to calculate the final kinetic energy (KE2).

6. Set ΔKE = KE2 and solve for the tension (Ft).

Your final answer should match the one you calculated using Newton's laws and constant acceleration kinematics. This is because the work-energy theorem and Newton's laws are equivalent ways of solving problems involving forces and motion.

I hope this helps! Let me know if you have any further questions.

Any help would be appreciated.

Your solution using Newton's laws and constant acceleration kinematics is correct. However, let's also solve this problem using work and energy.

First, we need to identify all the forces acting on the block. These include the normal force (Fn), the weight of the block (mg), and the tension force from the cord (Ft). Since the plane is frictionless, we can ignore the force of friction.

We can also define the initial and final positions of the block as the top and bottom of the incline, respectively. This means that the initial and final velocities are both zero.

Next, we can use the work-energy theorem, which states that the net work done on an object is equal to its change in kinetic energy. In this case, the net work done on the block is the work done by the tension force, which we can calculate using the formula W = Fd cosθ, where θ is the angle between the force and the displacement.

Since the block is moving along the incline, the angle between the tension force and the displacement is 15 degrees. Therefore, the work done by the tension force is W = Ft * 8m * cos15 = 8Ft * cos15.

Now, we can equate this to the change in kinetic energy of the block. Initially, the block has no kinetic energy, so the change in kinetic energy is just the final kinetic energy, which we can calculate using the formula KE = 1/2mv^2. Since the final velocity is zero, the final kinetic energy is also zero.

Therefore, we have 8Ft * cos15 = 0. Solving for Ft, we get Ft = 0.

This may seem counterintuitive, but it makes sense when we think about it. The block is coming to rest at the bottom of the incline, which means that its final velocity is zero. This also means that the net work done on the block is also zero. In other words, the tension force is doing just enough work to cancel out the gravitational potential energy of the block, but not enough to give it any kinetic energy.

In conclusion, the constant tension in the cord is 0N, which also confirms our previous solution using Newton's laws and constant acceleration kinematics.

## 1. What is the tension in a cord?

The tension in a cord refers to the pulling force that is applied to the cord in order to keep it taut.

## 2. How is the tension of a cord calculated?

The tension of a cord can be calculated using the equation T = F * sin(theta), where T is the tension, F is the force applied to the cord, and theta is the angle between the cord and the direction of the force.

## 3. Can the tension in a cord be greater than the applied force?

Yes, the tension in a cord can be greater than the applied force. This is because the tension is not only determined by the force applied, but also by the angle at which the force is applied.

## 4. How does the tension in a cord affect its strength?

The tension in a cord directly affects its strength. A higher tension means a stronger and more rigid cord, while a lower tension means a weaker and more flexible cord.

## 5. How can I solve a homework problem involving tension of a cord?

To solve a homework problem involving tension of a cord, you can follow these steps: 1) Identify the given information, including the applied force, angle, and mass of the object attached to the cord. 2) Use the equation T = F * sin(theta) to calculate the tension in the cord. 3) Use this tension value to determine if the cord can support the given mass. 4) If necessary, use other equations, such as Newton's laws of motion, to solve the problem. 5) Always double check your calculations and make sure your units are consistent.

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