Two physics KE/PE questions(and possibly more, studying for a test)

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SUMMARY

This discussion focuses on solving two physics problems related to kinetic energy (KE), potential energy (PE), and work done. In the first problem, a carpenter's hammer transfers heat energy to a nail, with calculations indicating that 1.5J of work is done, leading to a heat transfer of 1.73J after accounting for KE and PE. The second problem involves calculating the work done in accelerating a 1200kg car up an 18m hill at a speed of 12m/s, with conflicting results between participants, highlighting the importance of including both gravitational potential energy and kinetic energy in the calculations.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with the concepts of kinetic energy (KE) and potential energy (PE)
  • Basic knowledge of work-energy principles
  • Ability to perform unit conversions and basic algebra
NEXT STEPS
  • Review the work-energy theorem and its applications in physics
  • Study the relationship between force, distance, and work done in various contexts
  • Explore the concepts of heat transfer and energy conservation in mechanical systems
  • Practice solving problems involving KE and PE in different scenarios
USEFUL FOR

Students preparing for physics exams, educators teaching mechanics, and anyone interested in understanding the principles of energy transfer and work in physical systems.

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Homework Statement



a carpenter moves his .5 kg hammer at 3m/s when it contacts a nail. The nail receives an average force of 75N as it is driven 2cm int the wood board. The carpenter notices that the nail feels warm after he hits it with the hammer. How much heat energy was transferred to the nail.

Homework Equations


W = fd

The Attempt at a Solution



work w =force Fx distsnce D
w=75nx.02m
w=1.5j

Maybe I am supposed to do Ke = .5mv^2 and get 2.25J, then 2.25- 1.5J = .75J?

Except that was wrong?
Next problem

How much work is done by accelerating a 1200kg card from rest to a speed of 12m/s while it is movin up a frictionless hill that is 18m high.

force = mass x acceleration F= 1200x12=14400n
w=force x distance w=14400x18=259200j
=259.2kj

Except my friend is getting 298080J. Not sure whos right here. Thanks everyone!

Hes doing
W = mg(18) + .5m(12)^2
 
Last edited:
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When you calculate the work, you do need to take into account any change in potential energy. Try that for the first problem and see if it gets you a better result.
 
So for the first one, I tried

5mv^2 + mgh = W + Q
W = fd = 75N * .02m
= 1.5J
.5(.5kg)(3m/s)^2 = 2.25J
= KE
.5kg * 9.8m/s^2 * .02m = PE = .98J
2.25+ .98 = 1.5 + Q

So is Q 1.73J?