Answer:
If the speed of the merry-go-round doubles, the force you will need to exert to hang on is 400 N.
Explanation:
Given;
initial force exerted to hang on, F₁ = 100 N
The force exerted on the merry-go-round in order to hang on must be an inward force known as centripetal force.
Centripetal force is given by;
[tex]F_c = \frac{mv^2}{r} \\\\keeping \ "m" \ and \ "r" \ constant, we \ will \ have \ the \ following \ equation;\\\\\frac{F_c_1}{v_1^2} = \frac{F_c_2}{v_2^2} \\\\F_c_2 = \frac{F_c_1*v_2^2}{v_1^2}\\\\when \ the \ speed\ doubles \ i.e, v_2 = 2v_1\\\\ F_c_2 = \frac{F_c_1*(2v_1)^2}{v_1^2}\\\\ F_c_2 = \frac{F_c_1*4v_1^2}{v_1^2}\\\\F_c_2 = F_c_1 *4\\\\F_c_2 = 4(F_c_1)\\\\F_c_2 = 4 (100 \ N)\\\\F_c_2 = 400 \ N[/tex]
Therefore, If the speed of the merry-go-round doubles, the force you will need to exert to hang on is 400 N.
