Question

$\red{\boxed{\underline{QUESTION :}}}$
a uniform chain of length 2m is kept on a table of 50cm hangs freely from the edge of the table .the total mass of the chain is 5kg . what is the work done in pulling the entire chain on the table.(g=10m/s)

(A) 7.2J (B) 3J (C) 4.6J (D) 120J​

Answer 1

$\huge\boxed{\fcolorbox{orange}{orange}{Answer}}$

First of all let assume mas of chain hanging down is "m" while it's length is  "l".

Total length of chain "L"  while total mass is "M".

So from the given data we know that M is equal to 4kg

total length=L=2m

l = 60 cm = 0.6 m   

  As we know that 1 meter=100cm

 Thus we know that

m= I/L * M

m=(0.6/2 )* 4= 1.2kg 

In order to findout the tension of strain the formula is =mg

                                                                                      =1.2*10 = 12 Newton

we know that the tension is equal to force required and 0.6 is actually the displacement because that part is hanging

So  also we know

work done= force*displacement 

now as we have readings so just put the value 

Work done = 12 * 0.6 = 7.2 J

So the work done in this case will be equal to  7.2 J.

Answer 2

Answer:

Hey!✌️

Mass per unit length of chain =M/L=4/2=2kg/m

Work done in pulling the chain of length dx=dW=mgx=(2dx)(10)x

Inregrating both sides with proper limits x=0m to x=0.6m, we get

⇒W=∫ 00.6(20x)dx

⇒W=10(0.62−02)=3.6J

Hence correct answer is option B

Hope it will be helpful ✌️

@BadBoy