(1/2) m v^2 = (1/2) k x^2
.5 (1600) = .5 k (.04)
k = 40,000 Newtons/meter
F = k x = 40,000 (.2) = 8000 N
A catapult used to hold a stone of mass 500g is extended by 20cm with applied force .f if the stone leaves with a velocity of 40m/s,what is the value of f?
Help plz
3 answers
0.5 x 0.5(1600) = 0.5 x k x 0.04
K=20,000
F=4,000
K=20,000
F=4,000
According to Newton's second law
F = ma
a = v^2/2d
a = 1,600/0.4 = 4,000
:. F = 0.5 × 4000 = 2,000N
F = ma
a = v^2/2d
a = 1,600/0.4 = 4,000
:. F = 0.5 × 4000 = 2,000N