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Two objects of masses m1 = 0.56 kg and m2 = 0.88 kg are placed on a horizontal
frictionless surface and a compressed spring of force constant k = 280 N/m is placed
between them as in Fig. a. Neglect the mass of the spring. The spring is not attached to
either object and is compressed a distance of 9.8 cm. If the objects are released from rest,
find the final velocity of each object?

Sagot :

onyebuchinnaji

The final velocity of the first object is 5.03 m/s and the final velocity the second object is 6.48 m/s

The given parameters;

mass of the first object, M1 = 0.56 kg

mass of the second object, M2 = 0.88 kg

spring constant, k = 280 N/m

displacement of the spring, x = 9.8 cm = 0.098 m

Apply work energy theorem to determine the distance of each block;

[tex]W_1 + W_2 = \frac{1}{2} kx^2\\\\F_1d_1 + F_2d_2 = \frac{1}{2} kx^2\\\\(0.56\times 9.8)d_1 + (0.88\times 9.8)d_2 = 0.5\times 280 \times (0.098)^2\\\\5.488d_1 + 8.624d_2 = 1.345 ----(1)[/tex]

The total distance moved by the two blocks;

[tex]d_1 + d_2 = 0.098\\\\d_1 = 0.098 - d_2[/tex]

substitute d1 into equation 1

[tex]5.488(0.098-d_2) + 8.624d_2 = 1.345\\\\0.538 - 5.488d_2 + 8.624d_2 = 1.345\\\\3.136d_2 = 0.807\\\\d_2 = \frac{0.807}{3.136} = 0.257 \ m\\\\d_1 = 0.098 - 0.257\\\\d_1 = -0.159 \ m\\\\d_1 = 0.159 \ m (opposite \ direction \ to \ d_2)[/tex]

The acceleration of each object is calculated as;

[tex]F= kx\\\\F_1 = kx_1\\\\m_1a_1 = kx_1\\\\a_1 = \frac{kx_1}{m_1} = \frac{280 \times 0.159}{0.56} = 79.5 \ m/s^2\\\\a_2 = \frac{kx_2}{m_2} = \frac{280 \times 0.257}{0.88} = 81.77 \ m/s^2[/tex]

The final velocity of each object is calculated as;

[tex]v_f^2 = v_0^2 + 2as\\\\vf_1^2 = 0 + 2(79.5)(0.159) \\\\vf_1^2 = 25.281 \\\\v_f_1 = \sqrt{25.281} \\\\v_f_1 = 5.03 \ m/s\\\\For \ the \ second \ object;\\\\vf_2^2 = 0 + 2(81.77)(0.257)\\\\vf_2^2 = 42.03 \\\\v_f = \sqrt{42.03} \\\\v_f = 6.48 \ m/s[/tex]

Thus, the final velocity of the first object is 5.03 m/s and the final velocity the second object is 6.48 m/s

Learn more here: https://brainly.com/question/10063455

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