A stone of mass 50g is accelerated from a catapult to a speed of 8.0 ms-1 from rest over a distance of 30 cm. What average force is applied by the rubber of the catapult ?

Answer 1

average speed = (8+0)/2 = 4 m/s
so
t = .30 meters/4 meters/s =.3/4 s

Force * time = change of momentum
or impulse

F(.3/4) = .5 * 8

F = (5/3)(32) = 53.3 Newtons

Answer 2

wrong the answer of the question is 5N

Answer 3

5N

Answer 4

U= 0m/s
V= 8m/s
M= 50g =0.05kg
S=30cm =0.3m
t=?
F=?
S= (u+v/2) ×t
O.3= (0+8/2)t
0.3=4t
t= 0.3/4
t=0.075sec

Ft=M(V-U)
0.075F= 0.05×(8-0)
F=0.4/0.075
F= 5.33N