Respuesta :
Answer:
The book's speed is [tex]v=3.16 ms^{-1}[/tex].
Explanation:
The expression of the elastic potential energy is as follows;
[tex]PE=\frac{1}{2}kx^{2}[/tex] .......... (1)
Here, PE is the elastic potential energy, k is the spring constant and x is the distance.
The expression for kinetic energy is as follows;
[tex]KE=\frac{1}{2}mv^{2}[/tex] ............ (2)
Here, m is the mass of the object and v is the speed.
According to the given problem, a student places her 500 g physics book on a frictionless table. She pushes the book against a spring, compressing the spring by 4.0 cm, then releases the book. The elastic potential energy is converted into kinetic energy.
From (1) and (2),
PE=KE
[tex]\frac{1}{2}kx^{2}=\frac{1}{2}mv^{2}[/tex]
[tex]kx^{2}=mv^{2}[/tex]
Rearrange the above expression to calculate the book's speed.
[tex]v=x\sqrt{\frac{k}{m}}[/tex]
Convert the distance x from cm to m.
x= 4 cm
x= 0.04 m
Convert the mass m from g to kg.
m= 500 g
m= 0.2 kg
Put x= 0.04 m, m= 0.2 kg and [tex] k=1250 Nm^{-1}[/tex] in the expression for book's speed.
[tex]v=(0.04)\sqrt{\frac{1250}{0.2}}[/tex]
[tex]v=3.16 ms^{-1}[/tex]
Therefore, the book's speed is [tex]v=3.16 ms^{-1}[/tex].
The speed of the book as it slides away from the spring is 2 m/s.
The Potential energy of the compressed spring is also the energy needed for the book to slide away.
Using,
ke²/2 = mv²/2
ke² = mv²................. equation 1
Where k = spring constant, e = compression, m = mass of the book, v = speed of the book at which it slides.
make v the subject of the equation
v = √(ke²/m)........... Equation 2
From the question,
Given: k = 1250 N/m, e = 4 cm = 0.04 m, m = 500 g = 0.5 kg
Substitute these values into equation 2
v = √(1250×0.04²/0.5)
v = √(4)
v = 2 m/s
Hence, The speed of the book as it slides away from the spring is 2 m/s.
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