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Grade 12th passMechanics

A block of mass 10kg is placed over another block of mass 20kg which is placed on a smooth table. A constant horizontal force of 60 N is applied on 20kg block. Coefficient of friction between blocks is 0.5.find the net work done on the system in 2sec. ... .............. . Answer:240J

Profile image of Pawan joshi
7 Years agoGrade 12th pass
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1 Answer

Profile image of Agrata Singh
7 Years ago
Step1: Assume that the blocks move together with common acceleration a=F/(m1+m2)
a=60/(10+20)=2 m/s2
 
Step2:Find the max possible friction between the blocks
f(max)=uN=0.5xm1g=0.5x10x10= 50 
Accn due to this f(max), say af , is f/m1= 50/10=5m/s2
(Since friction provides forward accn to block 1 on the top,we only consider its mass for finding accn)
 
Now since af>a relative slipping is not possible
Hence the blocks move together with a=2m/s2
 
Displacement in 2 sec: 1/2at2=1/2x2x22=4m
 
W=Fd=60x4=240 J

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