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A BLOCK OF MASS 2kg. PLACED ON A LONG HORIZONTAL FRICTIONLESS TABLE IS PULLED HORIZONTALLY BY A CONSTANT FORCE F.IT IS FOUND TO MOVE 10m. IN FIRST TWO SECONDS. FIND THE MAGNITUDE OF F?

A BLOCK OF MASS 2kg. PLACED ON A LONG HORIZONTAL FRICTIONLESS TABLE IS PULLED HORIZONTALLY BY A CONSTANT FORCE F.IT IS FOUND TO MOVE 10m. IN FIRST TWO SECONDS. FIND THE MAGNITUDE OF F?

Grade:11

2 Answers

Menka Malguri
39 Points
9 years ago

since the table is frictionless,there is no opposing frictional force acting upon the block during its motion.

u=0 and t=2s and accn. of block a=F/m=F/2

then,

s=ut+1/2 at2

10=(0)2 +1/2 (F/2)(2)2

F=10N.

Gaurav Yadav
35 Points
9 years ago

Force = f

Mass = 2kg

Acc. = force/mass = f/2

s = ut + 1/2 X at X t

so 10 = 0 + 1/2 X f/2 X 4

so Force = 10N

 

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