Question

A 3.0 kg box is sliding along a frictionless horizontal surface with a speed of 1.8 m/s when it encounters a spring. (a) Determine the force constant (in N/m) of the spring, if the box compresses the spring 6.6 cm before coming to rest.
(b) Determine the initial speed (in m/s) the box would need in order to compress the spring by 1.8 cm.

Answer 1

Answer:

a) k = 2231.40 N/m

b) v = 0.491 m/s

Explanation:

Let k be the spring force constant , x be the compression displacement of the spring and v be the speed of the box.

when the box encounters the spring, all the energy of the box is kinetic energy:

the energy relationship between the box and the spring is given by:

1/2(m)×(v^2) = 1/2(k)×(x^2)

    (m)×(v^2) = (k)×(x^2)

a) (m)×(v^2) = (k)×(x^2)

                 k = [(m)×(v^2)]/(x^2)

                 k = [(3)×((1.8)^2)]/((6.6×10^-2)^2)

                 k = 2231.40 N/m

Therefore, the force spring constant is 2231.40 N/m

b) (m)×(v^2) = (k)×(x^2)

             v^2 = [(k)(x^2)]/m

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

                 v = \sqrt{ [(2231.40)((1.8×10^-2)^2)]/(3)}

                    = 0.491 m/s

Answer 2

The force constant of the spring compressed by the box is 2,231.4 N/m.

The initial speed of the box that will compress the spring to 1.8 cm is 0.49 m/s.

The given parameters;

• mass of the box, m = 3.0 kg
• speed of the box, v = 1.8 m/s
• extension of the spring, x = 6.6 cm = 0.066 m

The force constant of the spring is determined by applying the principle of conservation of energy;

$\frac{1}{2} kx^2 = \frac{1}{2} mv^2\\\\kx^2 = mv^2\\\\k = \frac{mv^2}{x^2} \\\\k = \frac{3 \times 1.8^2}{(0.066)^2} = 2,231.4 \ N/m$

The initial speed of the box that will compress the spring to 1.8 cm;

$\frac{1}{2} mv^2 = \frac{1}{2} kx^2\\\\mv^2 = kx^2\\\\v^2 = \frac{kx^2}{m} \\\\v = \sqrt{\frac{kx^2}{m}} \\\\v = \sqrt{\frac{2,231.4 \times (0.018)^2}{3}}\\\\v = 0.49 \ m/s$

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