Question

The 1.0-kg collar slides freely on the fixed circular rod. Calculate the velocity v of the collar as it hits the stop at B if it is elevated from rest at A by the action of the constant 58-N force in the cord. The cord is guided by the small fixed pulleys.

Answer 1

6.21 m/s

Step-by-step explanation:

Using work energy equation then

`U_(1-2)=T_B- T_A\\58d-mgh=0.5m(v_b^(2)-v_a^(2))`

where d is displacement from initial to final position, v is velocity and subscripts a and b are position A and B respectively, m is mass of collar, g is acceleration due to gravity

Substituting 1 Kg for m, 0.4m for h,
`v_a` as 0, 9.81 for g then

`58(\sqrt{0.4^(2)+0.3^(2)}-0.1)-(1* 9.81* 0.4)=0.5* 1* (v_b^(2)-v_a^(2))\\19.276=0.5* 1v_b^(2)\\v_b=6.209025688 m/s\approx 6.21 m/s`