m2*g-m1*g-F=ma/2
where
g(m2-m1)=ma
so
g(m2-m1)-F=g(m2-m1)/2
solve for F
Suppose that in the same Atwood setup another string is attached to the bottom of m1 and a constant force f is applied, retarding the upward motion of m1. If m1 = 5.30 kg and m2 = 10.60 kg, what value of f will reduce the acceleration of the system by 50%?
thank you!
2 answers
F= {9.8m/s^2 (10.6-5.30)}/ 2
F= {9.8 (5.3)}/2
F= 51.94/2
F= 25.97
{Significant figures} F = 26N
F= {9.8 (5.3)}/2
F= 51.94/2
F= 25.97
{Significant figures} F = 26N
1 answer