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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 ll 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

Suppose mm be the mass of the rope per unit length. Therefore, total mass >> of length L=mLL=m L \therefore Force applied F=(mL)aF=(m L) a or a=FmLa=\frac{F}{m L} So, tension in the rope at a distance ll == Force on the length (L0)(L-0) or T=m(Ll)a=m(1l)FmLT=m(L-l) a=m(1-l) \frac{F}{m L} or T=(1lL)FT=\left(1-\frac{l}{L}\right) F