Two boxes are initially at rest, side-by-side, on a floor with negligible fricti

Question

Two boxes are initially at rest, side-by-side, on a floor with negligible friction. Box A
has a mass of 2.0 kg, and box B has a mass of 6.0 kg. A worker pushes on box A with a
horizontal force P of 24 N. The boxes move (together) a distance of 3.0 m.

7. Identify the forces that do work on box A and calculate their magnitudes.
8. Identify the forces that do work on box B and calculate their magnitudes.
9. Identify the forces on A and B that do no work.
10. Find the net force on A and the net force on B.
11. How does the net work done on box A compare with the net work done on box B?
12. After the boxes have moved 3 m, how does the kinetic energy of box A compare
with the kinetic energy of box B?
13. How much work is done by the worker in pushing the boxes for 3 m?
14. What is the final total kinetic energy of the two boxes?

Explanation / Answer

the system acceleration rate: a=F/(m1+m2)=24/8=3(m/s2). 7. Identify the forces that do work on box A and calculate their magnitudes. F=24 N of the worker. F-F1=a*ma. so where F1 is the force exerted by box B on box A. F1=24-3*2=18(N) 8. Identify the forces that do work on box B and calculate their magnitudes. force exerted by box A on box B. F1'=18N 9. Identify the forces on A and B that do no work. normal force exerted by the floor and gravity force. 10. Find the net force on A and the net force on B. it is a*ma=6(N). a*mb=3*6=18(N) 11. How does the net work done on box A compare with the net work done on box B? three times. 12. After the boxes have moved 3 m, how does the kinetic energy of box A compare with the kinetic energy of box B? they both have the same velocity K=mv^2/2 so that Ka/Kb=ma/mb=1/3 13. How much work is done by the worker in pushing the boxes for 3 m? 24*3=72(J) 14. What is the final total kinetic energy of the two boxes? 72J

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