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Question

Physics Question on laws of motion

A uniform rope of length LL, resting on a frictionless horizontal surface, is pulled at one end by a force FF. The tension in the rope at a distance II from this end is

A

FF

B

lLF\frac{l}{L}F

C

LlF\frac{L}{l}F

D

(1lL)F\left( 1-\frac{l}{L} \right)F

Answer

(1lL)F\left( 1-\frac{l}{L} \right)F

Explanation

Solution

Mass of length l=mLl=m L, where mm is mass per unit length of the rope. From Newton's 2 nd law, Force == mass ×\times acceleration F=mLaF=m L a a=FmL\Rightarrow a=\frac{F}{m L} Tension in rope at distance ll from the given end T=m(Ll)aT =m(L-l) a =m(Ll)FmL=m(L-l) \frac{F}{m L} T=(1lL)FT=\left(1-\frac{l}{L}\right) F