Answer to Question #350253 in Physics for mlas2

Question #350253

A 1450-kg

 car is traveling with an initial speed of 24.0 m/s.



Determine the car's speed after −1.90  105 J

 of net work is done on the car.

 



Question 5.2b:

A 11.8-kg

 box is sliding across a horizontal floor. It has an initial speed of 1.45 m/s

 and the only force acting on it is kinetic friction with magnitude fk = 2.65 N.



Determine the distance the box will travel before coming to rest.


1
Expert's answer
2022-06-13T08:22:56-0400

1. The energy-work theorem says

"\\frac{mv_f^2}{2}=\\frac{mv_i^2}{2}+W"

"v_f=\\sqrt{v_i^2+2W\/m}"

"v_f=\\sqrt{24.0^2+2*(-1.90*10^5)\/1450}=17.7\\:\\rm m\/s"

2. The energy-work theorem says

"\\frac{mv_f^2}{2}-\\frac{mv_i^2}{2}=W"

"0-\\frac{mv_i^2}{2}=-f_kd"

So

"d=\\frac{mv_i^2}{2f_k}=\\frac{11.8*1.45^2}{2*2.65}=4.68\\:\\rm m"


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS