A 5 kg mass lies on a horizontal table. It is connected by a rope and pulley to

Question

A 5 kg mass lies on a horizontal table. It is connected by a rope and pulley to a 3 kg mass which is hanging over the table.The friction and mass of the pulley can be neglected in this problem.

a) When the system is set into downward motion, it descends with a constant speed of 1.00 m/s. What is the coefficient of kinetic friction between m1 and the table?

b) In part a), for each of the following forces :-
(i) gravity, (ii) tension in the string (iii) friction and (iv) the normal force, how much work is done on block m1 in a 10.0 s time interval?

c) Repeat b) for m2

d) Find the total work done on each block.

e) ) If the mass of m1 is doubled and the system is again set into motion with a downward speed of 1.1 m/s, what is its acceleration (magnitude and direction)?

Explanation / Answer

a). they descend with constant speed, so net force on 5kg mass is zero as well as of 3kg mass. let tension of the string be T> T=3*g. and T=5*g*k where k is the kinetic friction. so 5gk=3g so k=3/5=0.6 b) i) gravity. A=0 ii)tension. A=(3g)*(10*1)=294 iii) friction A=-5*g*k*(10*1)=-294(J). iv) normal force dont do any work. c) i). A=(3g)*(10*1)=294(J). ii) tension. A=(-3g)*(10*1)=-294(J). -- d) each block's total work done is zero. e) the kinetic friction F=10*g*0.6=58.8(N). so acceleration rate. 3g-10g*0.6=(m1+m2)*a so a=-2.3(m/s2). its negative, so its go upward.

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