How Does Friction Affect Motion on an Inclined Plane?

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SUMMARY

The discussion focuses on the physics of a 5kg lead block sliding down a 10m inclined plane at a 40° angle, reaching a speed of 6 m/s at the bottom. The potential energy (PE) at the top is calculated as 315.07 J using the formula PE = mgh, where the height is determined to be 6.43 m. The kinetic energy (KE) at the bottom is calculated as 90 J using KE = 1/2(m)(v²). The work done by friction and the coefficient of friction are to be determined using the relationship between the change in kinetic energy and the distance moved by the block.

PREREQUISITES
  • Understanding of gravitational potential energy (PE = mgh)
  • Knowledge of kinetic energy (KE = 1/2(m)(v²))
  • Familiarity with the concept of friction and its coefficient (Ffr = μ * Fn)
  • Basic trigonometry for calculating height from the incline angle
NEXT STEPS
  • Calculate the work done by friction using the difference in kinetic energy and distance moved.
  • Determine the normal force acting on the block on the incline.
  • Find the coefficient of friction using the formula μ = Ffr / Fn.
  • Explore the effects of varying incline angles on friction and motion.
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and inclined planes, as well as educators seeking to explain the concepts of energy, friction, and motion dynamics.

mmalone11
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1. Homework Statement
The diagram shows a 5kg block of lead released from rest at the top of an incline. The block has a speed of 6 m/s when it reaches the bottom. The angle between the slope and the ground is 40° and the slope is 10 m long.
a) What is its PE at the top?
b) What is its KE at the bottom?
c) What is the work done by friction?
d) What must be the coefficient of friction?


2. Homework Equations
I am having trouble finding the friction. Once i find the friction, i know how to find the coefficient force of friciton using Ffr=mu*Fn.

3. The Attempt at a Solution
For part a first I found what the height was by doing 10(sin(40))=Height and got 6.43 m. Than i plugged that into PE=mgh... 5(9.8)(6.43) ... getting 315.07J ...
For part b I used KE=1/2(m)(v2) ... 1/2(5)(62) ... getting 90J ...
 
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What is the component of the weight normal to the surface? That will be fn.
What is the component of the weight along the surface?
In the absence of the friction, what will be the KE at the bottom?
Difference in the KE = fr*d, where d is the distance moved by the block.
 
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