Answer:
(A) The speed just as it left the ground is 30.25 m/s
(B) The maximum height of the rock is 46.69 m
Explanation:
Given;
weight of rock, w = mg = 20 N
speed of the rock at 14.8 m, u = 25 m/s
(a) Apply work energy theorem to find its speed just as it left the ground
work = Δ kinetic energy
F x d = ¹/₂mv² - ¹/₂mu²
mg x d = ¹/₂m(v² - u²)
g x d = ¹/₂(v² - u²)
gd = ¹/₂(v² - u²)
2gd = v² - u²
v² = 2gd + u²
v² = 2(9.8)(14.8) + (25)²
v² = 915.05
v = √915.05
v = 30.25 m/s
B) Use the work-energy theorem to find its maximum height
the initial velocity of the rock = 30.25 m/s
at maximum height, the final velocity = 0
- mg x H = ¹/₂mv² - ¹/₂mu²
- mg x H = ¹/₂m(0) - ¹/₂mu²
- mg x H = - ¹/₂mu²
2g x H = u²
H = u² / 2g
H = (30.25)² / 2(9.8)
H = 46.69 m
We can find the force by using the following formula;
N = ma + mg
Fa = ma = 76 x 1.2 = 91.2
Fg = mg = 76 x 9.8 = 744.8
N = 91.2 + 744.8 = 836
So, the force is 836 N.
Answer:
avriage force F = 2722.5 N
Explanation:
For this problem we can use Newton's second law, to calculate the average force and acceleration we can find it by kinematics.
vf² = v₀² - 2 ax
The final carriage speed is zero (vf = 0)
0 = v₀² - 2ax
a = v₀² / 2x
a = 1.1²/(2 0.200)
a = 3.025 m / s²
a = 3.0 m/s²
We calculate the average force
F = ma
F = 900 3,025
F = 2722.5 N
Answer:
what are you saying man. They do provide structure.
Explanation:
Answer:
Obviously the answer is Sun...
