Calculating Velocity: Roofer's Toolbox Sliding on a 36deg Slanted Roof

  • Thread starter rezal
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In summary, the conversation discusses a problem involving a roofer accidentally nudging his toolbox while working on a slanted roof. The problem asks for the speed of the toolbox as it reaches the edge of the roof, given the initial distance from the edge and the kinetic friction force. The conversation also provides equations and a solution attempt, with suggestions for gaining confidence in one's work.
  • #1
rezal
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Hello, I'm practicing by doing the problems in my textbook but I have no way of knowing if my answers to positive questions are correct or not so can someone tell me if I did this problem correctly?

1. Homework Statement

While a roofer is working on a roof that slants at 36deg above the horizontal, he accidentally nudges his 85.0-N toolbox, causing it to start sliding downward, starting from rest. If it starts 4.25 m from the lower edge of the roof, how fast will the toolbox be moving just as it reaches the edge of the roof if the kinetic friction force on it is 22.0 N?

Homework Equations


KE + U + W = KE + U sin(theta)=o/4.25

The Attempt at a Solution


U + W = KE o=4.25sin36=2.49
mgcos(theta)(h)+wsin(theta)(u_k)=KE
(85*cos(36)*2.49)+(85*sin(36)*22)=KE
KE=1270.38
v=sqrt((2*KE)/m)
V=17.12 m/s?
 
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  • #2
Welcome to PF;
If you don't know if the answer is correct or not, then you did not understand the work. You are, after all, training to be able to solve problems where nobody knows the answers so the sooner you start figuring out how to gain confidence in your own work the better.

The way to gain confidence is to write out your reasoning, and reality check the results. Here you write out, in words, what the energy transformations are.
Then write down an energy equation that says the same thing.
Then go through the factors in the energy equation and write them out in terms that you are given ...

It is also best practice to sub numbers in at the very end if at all possible - do the algebra first.

Off the first glance, you should take another look at the energy calculations ... 22N is the friction force, so you don't need the coefficient of kinetic friction to calculate it. You are given the weight, so you don't need mg in your equation. the distance to the edge reads to me like the distance along the roof ... stuff like that.
 

1. What is the difference between kinetic energy and potential energy?

Kinetic energy is the energy an object possesses due to its motion, while potential energy is the energy an object possesses due to its position or state.

2. How are kinetic energy and potential energy related?

Kinetic energy can be converted into potential energy and vice versa. For example, when an object is lifted, its potential energy increases, but when it is dropped, the potential energy is converted into kinetic energy as it falls.

3. What is the formula for calculating kinetic energy?

The formula for calculating kinetic energy is KE = 1/2 * m * v^2, where m is the mass of the object and v is its velocity.

4. How can potential energy be calculated?

The formula for calculating potential energy depends on the type of potential energy. For gravitational potential energy, the formula is PE = m * g * h, where m is the mass, g is the acceleration due to gravity, and h is the height. For elastic potential energy, the formula is PE = 1/2 * k * x^2, where k is the spring constant and x is the displacement from equilibrium.

5. Can an object have both kinetic and potential energy at the same time?

Yes, an object can have both kinetic and potential energy at the same time. For example, a ball rolling down a hill has both kinetic energy due to its motion and potential energy due to its position on the hill.

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