4:15 PM 28% Ooo AT&T; moodle2.lsu.edu Take Home Problem Set 6 Due Date: October

Question

4:15 PM 28% Ooo AT&T; moodle2.lsu.edu Take Home Problem Set 6 Due Date: October 17, 2014 by 3:00 pm in Nicholson 202 TOTAL 20 PTS) For a lecture demo in front of class, a little cart (mass 0,5 kg) travels on a long, straight, level air track. Shown here is a graph of velocity vs. time of the cartLet's define "righr to be the positive direction. +velocity +4 m/s +2 m/s time (s) 1 2 2 m/s 1. (5 pts.) Calculate the net force on the cart att 3 s. Pay attention to the direction and show all your work. 2. (5 pts.) What is the velocity at t 2.5 seconds? Be sure to explain your answer (ie. in which direction is the car moving and what is the magnitude of the velocity) 3. (5pts) For the entire period of time from t 0tot 5 see, what is the network done on the cart? Be sure to show your work. 4. (5pts) At t-5 seconds, the air for the air track is suddenly turned off, so kinetic friction alone) causes the cart to grind to a halt in another 0.5 sec. Let the mass of the track itself be 5.0 kg, ie. ten times the mass of the little cart. Calculate the coefficient of kinetic friction between the cart and the track. Be sure to show your work. The following notes apply to all of the take-home problems this semester: l. Print this problem sheet and useitas a cover when you tum i your wurk. Work these problems on your own paper. Use as many sheets as youncod 3. Staple all of your submitted pages together before turming them in Make sure yaucomplete the top part of the cover page 4. Work the problems parts in order. Eg. you mayaot use work fium part b)to cu plate part 5. Clearly and completely explain your reasoning 6. Present your work as nearly as possible. If necessary, revise and copy your work for

Explanation / Answer

1)acceleration at t=3 s is slope of graph at t=3

=>a=-4/1=-4 m/s^2 -ve sign means discelaration

Force=m*a=0.5 *-4=-2 N     -ve sign means force is opposite to direction of motion

2)slope of that line is -4

So equation of line is v=-4t+C

to find C ,at t=3 v=0

=>C=12

at t=2.5 s

v=-4*2.5+12

=2 m/s

3)Force is applied only from t=2 to 3.5 s

in remaining time velosity is constant.So Force on cart=0

we found Force=-2N

distance travelled in t=2 to 3.5 is

S=ut+1/2at^2

S=4*1.5-1/2*4*1.5^2 = 1.5 m

So work on cart =2*1.5=3 J

4)Initial kinetic energy of cart=1/2*0.5*(-2)^2

                                         =1 J

By conservation of energy

final KE of system = 1/2*(5.5)*v^2=1

v=sqrt(2/5.5)

using v=u+at

a=(sqrt(2/5.5)-2)/0.5

Friction force F'=-0.5*g*coefficeient of friction=-0.5*(sqrt(2/5.5)-2)/0.5
=>coefficeient of friction=-0.5*(sqrt(2/5.5)-2)/(0.5*0.5*9.8)=0.285

Related Questions

Navigate

• Browse (All)
• Browse 4
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.