As per the question the rope is inclined at an angle of 60 degree with the horizontal.
Let the tension produced in the rope is T.
The effective force which pulls the toy along horizontal direction is 5 N.
Resolving T into two components we get-
[tex][1]\ \ Tsin\theta[/tex] in vertical upward direction.
[tex][2]\ \ Tcos\theta[/tex] in horizontal direction.
The vertical component is balanced by the weight of the body i.e
[tex]Tsin\theta=mg[/tex]
Hence the toy moves due to the horizontal component.
[tex]Hence\ Tcos\theta=5 N[/tex]
[tex]Tcos60= 5N[/tex]
[tex]T= \frac{5}{cos60}[/tex]
[tex]= 5*2 N[/tex] [ [tex]cos60 = \frac{1}{2}[/tex] ]
[tex]= 10 N[/tex]
Hence the tension produced in the rope is 10 N.
