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
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
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
But let's get serious for a second. To find the average force applied by the rubber of the catapult, we can use the kinematic equation:
v^2 = u^2 + 2as
where:
v = final velocity (8.0 m/s)
u = initial velocity (0 m/s, since it starts from rest)
a = acceleration
s = distance (0.30 m)
Rearranging the equation to solve for acceleration (a):
a = (v^2 - u^2) / (2s)
Substituting the given values:
a = (8.0^2 - 0^2) / (2 * 0.30)
Now we can solve for acceleration:
a = 106.6 m/s^2
To find the average force applied by the rubber of the catapult, we can use Newton's second law:
F = ma
Substituting the values:
F = (0.05 kg) * (106.6 m/s^2)
Calculating the average force:
F = 5.33 N
So, the average force applied by the rubber of the catapult is approximately 5.33 Newtons.
First, let's convert the mass of the stone from grams to kilograms.
Mass (m) = 50 g = 50/1000 kg = 0.05 kg
Next, we need to find the acceleration of the stone. We can use the kinematic equation:
v^2 = u^2 + 2as
where
v = final velocity = 8.0 m/s (given)
u = initial velocity = 0 m/s (since the stone is initially at rest)
a = acceleration
s = distance = 30 cm = 30/100 m = 0.3 m
Rearranging the equation, we get:
a = (v^2 - u^2) / (2s)
= (8.0^2 - 0^2) / (2 * 0.3)
= 64 / 0.6
= 106.67 m/s^2
Now we can calculate the average force using Newton's second law:
F = m * a
= 0.05 * 106.67
= 5.3335 N
Therefore, the average force applied by the rubber of the catapult is approximately 5.3335 Newtons.