Let F be the magnitude of the frictional force. This force performs an amount of work W on the bullet such that
W = -Fx
where x is the distance over which F is acting. This is the only force acting on the bullet as it penetrates the tree. The work-energy theorem says the total work performed on a body is equal to the change in that body's kinetic energy, so we have
W = ∆K
-Fx = 0 - 1/2 mv²
where m is the body's mass and v is its speed.
Solve for F and plug in the given information:
F = mv²/(2x)
F = (0.00426 kg) (881 m/s)² / (2 (0.0444 m))
F = 37,234.8 N ≈ 37.2 kN
PE = 0.5 × k × x²
PE potential Energy
k spring constant
x stretch/compression of the spring
Answer:
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Explanation:
Potential energy is when a object is not in motion and the ball is sitting on a shelf not being thrown around or rolling. kinetic energy is when a obeject is in motion and moving.
Answer:
m 200 g , T 0.250 s,E 2.00 J
;
2 2 25.1 rad s
T 0.250
(a)
2 2
k m 0.200 kg 25.1 rad s 126 N m
(b)
2
2 2 2.00 0.178 mm 200 g , T 0.250 s,E 2.00 J
;
2 2 25.1 rad s
T 0.250
(a)
2 2
k m 0.200 kg 25.1 rad s 126 N m
(b)
2
2 2 2.00 0.178 m
Explanation:
That is a reason
