- #1

Diana Rico

- 2

- 0

horizontal by a force of magnitude 140 N that acts parallel to the ramp. The

coefficient of kinetic friction between the ramp and the incline is If the suit-case

travels 3.80 m along the ramp, calculate

(a) the work done on the suitcase by the force

(b) the work done on the suitcase by the gravitational force;

(c) the work done on the suitcase by the normal force;

(d) the work done on the suitcase by the friction force;

(e) the total work done on the suitcase.

(f) If the speed of the suitcase is zero at the bottom of the ramp, what is its speed

after it has traveled 3.80 m along the ramp?

I feel like this problem is so simple yet I cannot get it!

a.) I used W=Frcos(theta) or W=140(3.8)cos(25) but the answer is not correct. The solution says to make theta=90, does that mean that I should use sin in order for it to equal 1 and not 0? The answer is just W=Fr or 532J but why?

b.) I tried to use the same formula because it is the same type of problem but with the angle as 115 degrees and Force as 196 N. W=-315J, but what I do not understand is the proof of why it works which states

*"Alternatively, the component of w parallel to the incline is wsin25. This component is down the incline so its angle with s is ∅=180°. Wwsin(25)=(196 Nsin25 )(cos180 )(3.80 m)=-315 J. The other component of w, wcos25, is perpendicular to s and hence does no work. Thus Ww=Wwsin(25)=-315J which agrees with the above."*Can someone please explain?

c.) I understood W=0 because the normal force and r are perpendicular.

d.) I do not know how the formulas 'n=wcos25' or 'fk=μkn' were obtained to find the answer. Does the the n= have do with the italicized part in part b?

e.) To find the total work done Wt= Wa+Wb+Wc+Wd or Wt= 532J+-315J+0J+(W found in part d) but I do not have part d in order to finish the problem.

f.) For this problem Wt=Kf-Ki or Wt=.5mvf^2-.5mvi^2 and since vi=0 W=.5mvf^2 and v=sqrt(2Wt/20) so all I need is Wt

So really I need help with understanding a, b, and how to get the formula for d please.