A 0.500 kg block is sitting on a horizontal, frictionless surface. The block is connected to a horizontal spring with a force co
nstant of 124 N/m. The other end of the horizontal spring rests against a wall. When a 100.0 g arrow is fired into the wooden block (where it becomes embedded), the spring compresses by 25 cm.
a) What is the maximum potential energy of the spring?
(b) What was the speed of the arrow and block just after collision?
(c) What was the initial kinetic energy of the arrow?
(d) Explain any difference between (a) and (b).
a) What is the maximum potential energy of the spring?
(b) What was the speed of the arrow and block just after collision?
(c) What was the initial kinetic energy of the arrow?
(d) Explain any difference between (a) and (b).
1 answer:
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Answer:
Explanation:
a ) angular frequency ω =
k is spring constant and m is mass attached
ω =
= 3.6515 rad / s
frequency of oscillation n = 3.6515 / (2 x 3.14)
= .5814 s⁻¹
x = .1 mcos(ωt)
= .1 mcos(3.6515t)
b ) maximum speed = ωA , A is amplitude
= 3.6515 x .1
= .36515 m /s
36.515 cm /s
maximum acceleration = ω²A
= 3.6515² x .1
= 1.333 m / s²
c ) Kinetic energy at displacement x
= 1/2 m ω²( A²-x²)
potential energy =1/2 m ω²x²
so 1/2 m ω²( A²-x²) = 1/2 m ω²x²
A²-x² = x²
2x² = A²
x = A / √2
Kinetic energy is the energy of a moving object.
Kinetic energy can be calculated using the following equation;
K.E =
where - m - mass of object
v - speed of object
substituting the values in the equation
K.E = 1/2 x 21 kg x (1 m/s)²
K.E = 10.5 J
Kinetic energy can be calculated using the following equation;
K.E =
where - m - mass of object
v - speed of object
substituting the values in the equation
K.E = 1/2 x 21 kg x (1 m/s)²
K.E = 10.5 J