Grade 11Mechanics

Two blocks of different mass are hanging on two ends of a string passing over a friction less pulley.The heavier block has twice the mass as that of the lighter one.The tension in the string is of 60N.The decrease in potential energy during the first second after the system is released is 15k.Then find the value of k.

debadutta mishra

15 Years agoGrade 11

3 Answers

vikas askiitian expert

Approved Tutor Answer15 Years ago

let the mass of lighter and heavier block be m and 2m respectively....

for block m

T-mg=ma................1

for block 2m

2mg-T=2ma...............2

adding 1 and 2

3ma=mg and T=4ma..........1

accleration of both blocks is same so displament after 1 sec for both will be equal

d=ut+at^2/2 (u=0 and t=1s)

d=a/2

lighter block will be displaced upward and heavier block is displaced downward so total potential energy

stored is PE= 2mgd + mg(-d)

=mgd..............2

solving 1 and 2

PE=mga/2=mgT/8m

PE =Tg/8

change in PE is 15K given

15K=Tg/8

K=5

Gaudham

8 Years ago

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Rishu

7 Years ago

Given

a=1m/sec^2

t=4sec

v=u+at

v=4m/sec

ma=F-2gsin30

F=12N

P=dW/dt

P=F.dr/dt

P=F.v

P=12*4=48

Now

12K=48

K=4

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debadutta mishra

3 Answers

vikas askiitian expert

Gaudham

Rishu

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