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Question: A uniform thick rope of length \(5\,m\) is kept on a frictionless surface and a force of \(5\,N\) is...

A uniform thick rope of length 5m5\,m is kept on a frictionless surface and a force of 5N5\,N is applied to one of its ends. Find tension in the rope at 1m1\,m from this end.
A. 1N1\,N
B. 3N3\,N
C. 4N4\,N
D. 5N5\,N

Explanation

Solution

In order to solve this question, we will first calculate mass per unit length of the rope and the net acceleration produced on the rope due to applied force and then we will use the concept that at any point on the rope the tension force will be responsible for motion of remaining part of the rope to calculate tension force.

Formula used:
Force applied on a body is FF and mm is the mass of a body then acceleration aa is given by Newton’s second law of motion.
a=Fma = \dfrac{F}{m}

Complete step by step answer:
According to the question, we have given that
Total length of rope, l=5ml = 5m
Applied force on the rope, F=5NF = 5N
Let ‘m’ be the total mass of the rope then mass per unit length m’ will be
m=ml m=m5m' = \dfrac{m}{l} \\\ \Rightarrow m'= \dfrac{m}{5}
let a be the acceleration produced in the rope then using,
a=Fma = \dfrac{F}{m}
a=5m\Rightarrow a = \dfrac{5}{m}

Now, we need to calculate tension at a distance of 1m1m from the end of applied force, then this tension force T will produce acceleration of remaining part of rope which will be of length l1=51=4ml - 1 = 5 - 1 = 4m
So, tension force T will move the net mass of 4m4m length which can be calculated as
M=mlM = m'l' where M is the mass of 4m4m length and l=4l' = 4 put m=m5m' = \dfrac{m}{5} we get,
M=4m5\Rightarrow M = \dfrac{{4m}}{5}

Now tension force TT will produce same acceleration of this part of rope as same as of complete rope which we have already calculated as a=5ma = \dfrac{5}{m} so we have,
T=MaT = Ma
On putting values we get,
T=4m5×5mT = \dfrac{{4m}}{5} \times \dfrac{5}{m}
T=4N\therefore T = 4N

Hence, the correct option is C.

Note: It should be remembered that, mass of any part of the rope is simply the product of length of that part and mass per unit length of the rope and here tension force at any point on the rope will be of different value but the sum of tension forces due to two parts of rope will always be equal to net force applied on the rope.