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A massless spring with force constant 602 N/m is fastened at its left end to a vertical wall, as shown below. Initially, the 6 kg block and 3 kg block rest on a horizontal surface with the 6 kg block in contact with the spring (but not compressing it) and with the 3 kg block in contact with the 6 kg block. The 6 kg block is then moved to the left, compressing the spring a distance of 0.6 m, and held in place while the 3 kg block remains at rest as shown below. The acceleration of gravity is 9.8 m/s2

A) Determine the elastic energy U stored in the compressed spring. Answer in units of J ?

B) The 6 kg block is then released and accelerates to the right, toward the 3 kg block. The surface is rough and the coefficient of friction between each block and the surface is 0.3 . The two blocks collide, stick together, and move to the right. Remember that the spring is not attached to the 6 kg block. Find the speed of the 6 kg block just before it collides with the 3 kg block. ? Answer in units of m/s.

C) Find the final speed of both blocks (stuck together) just after they collide. Answer in units of m/s?

D) Find the horizontal distance the blocks move before coming to rest. Answer in units of m ?

Respuesta :

snehashish65

(A) The elastic energy stored in the compressed spring is 108.36 J.

(B) The speed of the 6 kg block just before it collides with the 3 kg block is   6.29 m/s.

(C) The speed of each block after the collision is 4.19 m/s.

(D) The distance covered by each block before coming to rest is 2.98 m

Given data:

Value of force constant is, [tex]k=602 \;\rm N/m[/tex].

Mass of block 1 is, [tex]m_{1}=6 \;\rm kg[/tex].

Mass of block 2 is, [tex]m_{2}=3 \;\rm kg[/tex].

Spring compression distance is, [tex]x=0.6 \;\rm m[/tex].

Gravitational acceleration is, [tex]g = 9.8 \;\rm m/s^{2}[/tex].

Coefficient of friction between each block and the surface is,[tex]\mu=0.3[/tex].

(A)

The elastic energy (U) stored in the compressed spring is,

[tex]U=\dfrac{1}{2}kx^{2}\\U=\dfrac{1}{2}\times 602\times 0.6^{2}\\U=108.36 \;\rm J[/tex]

Thus, the elastic energy stored in the compressed spring is 108.36 J.

(B)

Apply the Work-energy concept as,

Work done by frictional force = Kinetic energy + Elastic energy

[tex]W =KE -U\\f \times x= \dfrac{1}{2}m_{1}v^{2}-U\\(\mu \times m_{1} \times g) \times x= \dfrac{1}{2}m_{1}v^{2}-U \\(0.3 \times 6 \times 9.8) \times 0.6= \dfrac{1}{2} \times 6 \times v^{2}-108.36\\\\v^{2}=\dfrac{(0.3 \times 6 \times 9.8 \times 0.6)+108.36}{3}\\v=6.29 \;\rm m/s[/tex]

Thus, the speed of the 6 kg block just before it collides with the 3 kg block is 6.29 m/s.

(C)

Apply the conservation of linear momentum as,

Initial momentum = Final momentum

[tex]m_{1}v+m_{2}v'=(m_{1}+m_{2})V\\[/tex]

Here, v' is the initial speed of 3 kg block, which was at rest. So, v'=0.

And V is the final speed of both blocks after the collision.

Solving as,

[tex]6 \times 6.29 +3 \times 0=(6+3)V\\V = \dfrac{ 6 \times 6.29}{9} \\V=4.19 \;\rm m/s[/tex]

Thus, the speed of each block after the collision is 4.19 m/s.

(D)

Again use the Work energy theorem to obtain the distance (s) covered by each block before coming to rest as,

Work done by frictional force = Kinetic energy of both blocks

[tex]W = KE_{net}\\f \times s = \dfrac{1}{2}(m_{1}+m_{2})V^{2}\\\mu \times (m_{1}+m_{2}) \times g \times s = \dfrac{1}{2}(m_{1}+m_{2})V^{2} \\\mu \times g \times s = \dfrac{1}{2} \times V^{2}\\0.3 \times 9.8 \times s = \dfrac{1}{2} \times 4,19^{2}\\s = 2.98 \;\rm m[/tex]

Thus, the distance covered by each block before coming to rest is 2.98 m.

Learn more about kinematics of motion here:

https://brainly.com/question/19818185?referrer=searchResults

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