Question

Determine the final velocity of 15-kg mass moving horizontally if it starts at 10m/s moves a distance of 10m while the following net force acts in the direction of motion (where s is the distance in the direction of motion) :
A) 200N
B)20s N
C)200 cos s phi/20 N

Answer 1

I think I can use the kinematic equations for question A) and I know that I have to use integral for B) and C). But, how do I use the integrals?

Answer 2

\[F=m·a=m \frac{ dv }{ dt }\rightarrow dW=F·ds=m·\frac{ dv }{ dt }ds=m·v·dv\]Then:\[\int\limits\limits_{S_A}^{S_B}F·ds=\int\limits\limits_{v_A}^{v_B}m·v·dv=\frac{ 1 }{ 2 }·m(v_B^2-v_A^2) \rightarrow v_B^2=v_A^2+\frac{ 2 }{ m }\int\limits_{S_A}^{S_B}F·ds\]Then for case B:
\[v_B^2=100+\frac{ 2 }{ 15 }\int\limits_{0}^{10}20s·ds=100+\frac{ 4 }{ 3 }·100=700/3\rightarrow v_B=15.28\]

Answer 3

I still don't understand how \[dW=m \frac{ dv }{ dt }ds\] turns into\[dW=m.v.dv\]

btw, is it possible for me to do this :\[W=\int\limits_{0}^{10}20s.ds=2000 Joule\]\[\frac{ W }{ s }= Favg \rightarrow Favg=\frac{ 2000 }{ 10 }=200N\]

Answer 4

Just move a little the dt in the denominator right under the ds. Now you have ds/dt=v

Answer 5

What the equation is telling you is that the work done by the force turns into a variation of kinetic energy

Answer 6

Okay, I think I get it.. Thank you very much for your help.. :)

Answer 7

You are welcome. By the way review the integral you have solved, it is 1000Joule and not 2000