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
Answer:
9758 how many significant figures
Answer:
0.0312J
Explanation:
Let x be the distance the staple moves:
And spring constant is 
Hence, the potential energy is 0.0312J
