Physics Work Energy Theorem Question?

Question

A 15.0 kg cart is moving with a velocity of 7.50 m/s down a level hallway. A constant force of -10.0 N acts on the cart, and its velocity becomes 3.20 m/s.

What is the change in kinetic energy of the cart?

How much work was done on the cart?

How far did the cart move while the force acted?

Answer 1

> "What is the change in kinetic energy of the cart?"



The kinetic energy always equals ½mv²



Initial KE = ½m(7.50 meters/s)²

Final KE = ½m(3.20 meters/s)²



So the CHANGE is...well, how much it changed. That is, the final KE minus the initial KE



> "How much work was done on the cart?"



The Work/Energy theorem says (memorize this): "The change in an object's KE equals the total work done on the object by all forces." So, you just calculated the change in KE. So that's the amount of work done.



> "How far did the cart move ..."



Work = force × (distance moved)



Through the magic of algebra:



distance moved = force / work



They give you the force, and you just calculated the work. Divide.

Answer 2

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Answer 3

W = ΔK = 1/2m[(v₂)² - (v₁)²]



ΔK; change in kinetic energy of the cart

W = F*d; work = force time distance(displacement)





v₁= 7.50 m/s

v₂= 3.20 m/s

F = 3.20 m/s

m = 15.0 kg



so



ΔK = 1/2*15.0 kg*[(3.20 m/s)² - (7.50 m/s)²] = -345 J



W = 3.20 m/s * ({1/2*15.0 kg*[(3.20 m/s)² - (7.50 m/s)²]}/ 3.20 m/s ) = ΔK



d = {1/2*15.0 kg*[(3.20 m/s)² - (7.50 m/s)²]}/ -10.0 N = 3.45 m

Answer 4

delta KE = 1/2m[V1² - V2²] = (0.5)(15)(7.5² - 3.2²) = 7.5(46.0) = 345 J ANS-1

ON cart = work is negative as force is opposite cart's motion = -345 J ANS-2

Work done/Force applied = Distance moved <= U can solve this last part, Ok

Answer 5

A) delta KE = 0.5* 15kg*(7.5 - 3.2)^2



B) Same as A



C) Work = force * distance = change in KE



Work = change in KE = 10 N * distance