A) How fast was the brick moving after 4s?
Vf=?
Vi=0 (because it was dropped, not thrown)
A= -9.8m/s^2 (gravity)
t= 4s
Use the equation Vf=Vi+A(t)
Vf=0+(-9.8)(4)
Final answer: Vf= -39.2m/s
b) How far did the brick fall after 4s?
D=?
Vi=0
t=4s
A=-9.8m/s2
**You do have the final velocity, but it is best to avoid using numbers that you have calculated yourself.**
Use the equation: d=Vi(t)+0.5(A)(t)^2
d=(0)(4)+0.5(-9.8)(4)^2
d=(-4.9)(16)
d=-78.4m
Therefore, after 4s the brick fell 78.4m
Answer:
a) a = - 0.0833 m / s², b) t = 4.4 s
Explanation:
a) this is a kinematics exercise where the acceleration is along the inclined plane
v = v₀ - a t
a = v₀ - v / t
a = 3 - 8/60
a = - 0.0833 m / s²
b) in this case the final velocity is zero
v = v₀ - a t
0 = v₀ - at
t = v₀ / a
t = 28 / 6.4
t = 4.375 s
t = 4.4 s
Explanation:
It is given that,
Mass of the brick, m = 1.15 kg
Radius of the circle, r = 1.44 m
The cable will break if the tension exceeds 43.0 N
Let v is the maximum sped can have at the bottom of the circle before the cable will break. At the bottom of the circle, the net force is equal to the centripetal force along with the weight of the brick. So,
v = 6.30 m/s
So, the maximum speed of the brick at the bottom of the circle before the cable will break is 6.3 m/s. Hence, this is the required solution.
