Respuesta :
Answer:
(a)
[tex]0.063[/tex] m
(b)
[tex]0.116 [/tex] m
Explanation:
(a)
[tex]U[/tex] = Potential energy stored in the spring = 3.20 J
[tex]k[/tex] = Spring constant of the spring = 1600 N m⁻¹
[tex]x[/tex] = Compression of the spring
Potential energy stored in the spring is given as
[tex]U = (0.5) kx^{2}[/tex]
inserting the values
[tex]3.20 = (0.5) (1600)x^{2}[/tex]
[tex]3.20 = (800) x^{2}[/tex]
[tex]x^{2}=\frac{3.20}{800}[/tex]
[tex]x=\sqrt{\frac{3.20}{800}} = 0.063[/tex] m
b)
[tex]m[/tex] = mass of the book being dropped = 1.20 kg
[tex]h[/tex] = height of the book being dropped above the top of spring = 0.80 m
[tex]d[/tex] = distance by which the spring gets compressed
Assuming the reference point for measuring the gravitational potential energy to be as the lowest point of the book after compression of the book.
Using conservation of energy
Spring Potential energy = Potential energy of the book
[tex](0.5) kd^{2} = mg(h + d)[/tex]
[tex](0.5) (1600)d^{2} = (1.20)(9.8)(0.80 + d)[/tex]
[tex](800) d^{2} = 9.408 + 11.76 d[/tex]
Solving the quadratic equation, we get
[tex]d = 0.116 [/tex] m
The maximum distance by which the spring will be compressed is 0.116m.
What is the spring Potential energy?
The spring potential energy is the energy stored in the spring due to its compression. It is given by the formula,
[tex]U = \dfrac{1}{2}kx^2[/tex]
A.) We know that the potential energy stored in the spring is given by the formula,
[tex]U = \dfrac{1}{2}kx^2[/tex]
As it is given that the energy needed to be stored in the spring is 3.20 J, while the force constant is 1600 N/m. therefore, the potential energy can be written,
[tex]3.20 = \dfrac{1}{2}\times 1600 \times x^2\\\\x = 0.063\rm\ m[/tex]
Thus, the distance that the spring must be compressed for 3.20 J of potential energy to be stored in it is 0.063m.
B.) As we know that the book is falling on the spring, therefore, the potential energy of the book will be transformed to the potential energy of the spring.
[tex]\dfrac{1}{2}kx^2 = mgh[/tex]
Since the book is placed 0.80 m above the top of the spring, therefore, the height of the book is the sum of the height of the book and the height of the spring.
[tex]\dfrac{1}{2}\times 1600 \times x^2 = 1.20 \times 9.81 \times (x + 0.80)\\\\800x^2 = 9.408 + 11.76x\\\\800x^2-11.76 x -9.408=0[/tex]
Solving the quadratic equation,
x = 0.116 m
Hence, the maximum distance by which the spring will be compressed is 0.116m.
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