Question: Use work and energy to find Paul's speed after being pulled 2.60 m.
Answer:
Paul's speed is 2.93m/s.
Explanation:
Let's call the tension on the rope [tex]T[/tex] and [tex]M[/tex] Paul's mass.
The normal force [tex]N[/tex] on Paul equals his weight [tex]Mg[/tex] minus the upward vertical pull from the rope [tex]T_y[/tex]:
[tex]N = Mg- T_y[/tex]
since [tex]T_y = T sin(\theta)[/tex]
[tex]N = Mg- Tsin(\theta)[/tex]
Therefore, the frictional force is
[tex]F_s = \mu N = \mu (Mg-Tsin(\theta))[/tex].
Thus, the net horizontal force on Paul is
[tex]F_{net} = T_x-F_s[/tex]
where [tex]T_x[/tex] is the horizontal component of rope's tension and it is
[tex]T_x = T cos(\theta)[/tex]
therefore, the net force becomes
[tex]F_{net} = Tcos(\theta)-\mu(Mg-Tsin(\theta))[/tex]
Now, when this horizontal net force pulls Paul by a horizontal distance [tex]d[/tex], the work done is
[tex]W = (F_{net})d=d(Tcos(\theta)-\mu(Mg-Tsin(\theta)))[/tex]
and according to the work-kinetic energy theorem
[tex](F_{net})d = \dfrac{1}{2}Mv^2[/tex]
which gives
[tex]d(Tcos(\theta)-\mu(Mg-Tsin(\theta)))=\dfrac{1}{2}Mv^2[/tex]
solving for [tex]v[/tex] we get:
[tex]v=\sqrt{\dfrac{2d(Tcos(\theta)-\mu(Mg-Tsin(\theta)))}{M} }[/tex]
Finally, we put in numerical values
[tex]T = 29.0N[/tex],
[tex]M= 30kg[/tex],
[tex]\theta =30^o[/tex]
[tex]\mu =0.170[/tex],
[tex]d =2.60m[/tex]
to get
[tex]v=\sqrt{\dfrac{2(2.60m)((29.0N)cos(30^o)-(0.170)((13.0kg)(9.8m/s^2)-(29.0N)sin(30^o)))}{13.0kg} }[/tex]
[tex]\boxed{v = 2.93m/s.}[/tex]
Hence, when Paul is pulled by 2.60m, his speed becomes 2.93m/s.
