Suppose a 4.0-kg projectile is launched vertically with a speed of 8.0 m/s. What is the maximum height the projectile reaches?
1 answer:
Answer:
h = 3.3 m (Look at the explanation below, please)
Explanation:
This question has to do with kinetic and potential energy. At the beginning (time of launch), there is no potential energy- we assume it starts from the ground. There, is, however, kinetic energy
Kinetic energy = m
Plug in the numbers = (4.0)(
)
Solve = 2(64) = 128 J
Now, since we know that the mechanical energy of a system always remains constant in the absence of outside forces (there is no outside force here), we can deduce that the kinetic energy at the bottom is equal to the potential energy at the top. Look at the diagram I have attached.
Potential energy = mgh = (4.0)(9.8)(h) = 39.2(h)
Kinetic energy = Potential Energy
128 J = 39.2h
h = 3.26 m
h= 3.3 m (because of significant figures)
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Use the kinematics formula: y= yo+Vo-1/2at^2
Yo- initial position
Vo- initial velocity
Yo- initial position
Vo- initial velocity
Answer:
A) 0.188
B) 53.1 ⁰
Explanation:
taking moment about 0
∑ Mo = Lo∝
mg 1/2 sin∅ = 1/3 m L^2∝
note ∝ = w∅
forces acting along t-direction ( ASSUMED t direction)
∑ Ft = Ma(t) = mr∝
mg sin ∅ - F = m* 1/2 * 3g/2l sin∅
therefore F = mg/4 sin∅
forces acting along n - direction ( ASSUMED n direction)
∑ Fn = ma(n) = mr()
= mg cos∅ - N = m*1/2*3g/1 ( 1 - cos∅ )
hence N = mg/2 ( 5cos∅ -3 )
A ) Angle given = 30⁰c find coefficient of static friction
∪ = F/N
= = 0.188
B) when there is no slip
N = O
= 5 cos ∅ -3 =0
therefore cos ∅ = 3/5 hence ∅ = 53.1⁰
Answer:
the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB = 0 is 31.3 m/s
Explanation:
given information
car's mass, m = 1200 kg
= 100 m
=
= 150 m
= 0
according to conservative energy
the distance from point A to B, h = 150 m - 100 m = 50 m
the initial speed
final speed = 0
thus,
² =
² - 2 g h
0 = ² - 2 g h
² = 2 g h
= √2 g h
= √2 (9.8) (50)
= 31.3 m/s