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horizontal block–spring system with the block on a frictionless surface has total mechanical energy E 5 47.0 J and a maximum displacement from equilibrium of 0.240 m. (a) What is the spring constant? (b) What is the kinetic energy of the system at the equilibrium point? (c) If the maximum speed of the block is 3.45 m/s, what is its mass? (d) What is the speed of the block when its displacement is 0.160 m? (e) Find the kinetic energy of the block at x 5 0.160 m. (f) Find the potential energy stored in the spring when x 5 0.160 m. (g) Suppose the same system is released from rest at x 5 0.240 m on a rough surface so that it loses

Respuesta :

Answer:

Part a)

[tex]k = 1632 J[/tex]

Part b)

[tex]KE = 47 J[/tex]

Part c)

[tex]m = 7.9 kg[/tex]

Part d)

[tex]v = 2.57 m/s[/tex]

Part e)

[tex]KE = 26.1 J[/tex]

Part f)

[tex]PE = 20.9 J[/tex]

Explanation:

Total Mechanical energy is given as

[tex]E = 47.0 J[/tex]

Its maximum displacement from mean position is given as

[tex]A = 0.240 m[/tex]

Part a)

Now from the formula of energy we know that

[tex]E = \frac{1}{2}kA^2[/tex]

[tex]47 = \frac{1}{2}k(0.240)^2[/tex]

[tex]k = 1632 J[/tex]

Part b)

At the mean position of SHM whole mechanical energy will convert into kinetic energy

so it is given as

[tex]KE = 47 J[/tex]

Part c)

As per the formula of kinetic energy we know that

[tex]KE = \frac{1}{2}mv^2[/tex]

[tex]47 = \frac{1}{2}m(3.45^2)[/tex]

[tex]m = 7.9 kg[/tex]

Part d)

As we know by the equation of the speed of SHM is given as

[tex]v = \sqrt{\frac{k}{m}(A^2 - x^2)}[/tex]

[tex]v = \sqrt{\frac{1632}{7.9}(0.24^2 - 0.16^2)}[/tex]

[tex]v = 2.57 m/s[/tex]

Part e)

As we know by the formula of kinetic energy

[tex]KE = \frac{1}{2}mv^2[/tex]

[tex]KE = \frac{1}{2}(7.9)(2.57^2)[/tex]

[tex]KE = 26.1 J[/tex]

Part f)

As per energy conservation we know

KE + PE = Total energy

[tex]26.1 + PE = 47[/tex]

[tex]PE = 20.9 J[/tex]

shubhamchouhanvrVT

(a) The spring constant will be k=1632 [tex]\frac{N}{m^2}[/tex]

(b) The KE at equilibrium point =47j

(c) The mass =7.9kg

(d) The velocity block [tex]2.57\frac{m}{sec}[/tex]

(e)The KE of block 0.160m will be 26.1

(f) The PE will be 20.9j

What will be the asked values of the spring-mass system in the question?

(a) for finding spring constant

KE=47 j

A=0.240m

By using the formula

[tex]E=\dfrac{1}{2} kx^2[/tex]

[tex]47=\dfrac{1}{2} k(0.240)^2[/tex]

[tex]k=1632\frac{N}{m^2}[/tex]

(b) At the mean position the whole mechanical energy will be equal to KE so

KE=47j

(c) The mass of the system

[tex]KE =\dfrac{1}{2} mv^2[/tex]

[tex]47=\dfrac{1}{2} m(3.45^2)[/tex]

[tex]m=7.9kg[/tex]

(d)Now the speed of the block

[tex]v=\sqrt{\dfrac{k}{m} (A^2-x^2)}[/tex]

[tex]v=\sqrt{\dfrac{1632}{7.9} (0.24^2-0.16^2)}[/tex]

[tex]v=2.57\frac{m}{s}[/tex]

(e) The KE of the block

[tex]KE=\dfrac{1}{2} mv^2=\dfrac{1}{2} 7.9(2.57)^2=26.1J[/tex]

(f) The PE of the system

[tex]Total Energy = KE+PE[/tex]

[tex]PE= 47-26.1 =20.9J[/tex]

Thus

(a) The spring constant will be k=1632 [tex]\frac{N}{m^2}[/tex]

(b) The KE at equilibrium point =47j

(c) The mass =7.9kg

(d) The velocity block [tex]2.57\frac{m}{sec}[/tex]

(e)The KE of block 0.160m will be 26.1

(f) The PE will be 20.9j

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