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

mvf22=mvi22+W\frac{mv_f^2}{2}=\frac{mv_i^2}{2}+W

vf=vi2+2W/mv_f=\sqrt{v_i^2+2W/m}

vf=24.02+2(1.90105)/1450=17.7m/sv_f=\sqrt{24.0^2+2*(-1.90*10^5)/1450}=17.7\:\rm m/s

2. The energy-work theorem says

mvf22mvi22=W\frac{mv_f^2}{2}-\frac{mv_i^2}{2}=W

0mvi22=fkd0-\frac{mv_i^2}{2}=-f_kd

So

d=mvi22fk=11.81.45222.65=4.68md=\frac{mv_i^2}{2f_k}=\frac{11.8*1.45^2}{2*2.65}=4.68\:\rm m


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