Question

A system of two blocks are connected by a string. These blocks are horizontally accelerated by an applied force of magnitude F on the frictionless horizontal surface. If the right block is 5kg, and the left block is 3kg, what is tension in the string between the blocks is:

(A) 3 F
(B) 5 F
(C) 3/8 F
(D) 1/3 F

Answer 1

The tension in the string between the blocks is: (C) 3/8 F

Step-by-step explanation:

We need to apply the Newton's second Law
`F=m*a`. First we need to add both block masses as:
`m_(T)=m_(r) +m_(l)=3+5=8(Kg)`. Then we can find the acceleration of the system as:
`a_(T)=(F)/(m_(T) )=(F)/(8)`. Therefore solving for T from the free body diagram, we find:
`F_(r) =m_(r)*a_(T)=(5*F)/(8)`, and adding forces on axle x we get:
`F_(r)= (5*F)/(8) =F-T`. Now we can calculate the tension in the string as:
`T=F-(5*F)/(8)=(3F)/(8)`.