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Question

A 263-gram block is dropped on a vertical spring with a constant force a = 2.52 N / cm (Figure 20). The block sticks to the resource, and the resource is compressed 11.8 cm before momentarily coming to rest. While the resource is being compressed, how much work is done on (a) the force of gravity and (b) the resource? (c) What was the speed of the block immediately before it reached the resource? (d) If this initial velocity of the block is doubled, what is the maximum compression of the spring?

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Answer to a math question A 263-gram block is dropped on a vertical spring with a constant force a = 2.52 N / cm (Figure 20). The block sticks to the resource, and the resource is compressed 11.8 cm before momentarily coming to rest. While the resource is being compressed, how much work is done on (a) the force of gravity and (b) the resource? (c) What was the speed of the block immediately before it reached the resource? (d) If this initial velocity of the block is doubled, what is the maximum compression of the spring?

Expert avatar
Miles
4.9
99 Answers
(a) \, T_g = m \times g \times h

T_g = 0.263 \, kg \times 9.8 \, m/s^2 \times 0.118 \, m = 0.304 \, J

(b) \, T_r = -\frac{1}{2} \times k \times x^2

k = \frac{2.52 \, N/cm}{0.01 \, m/cm} = 252 \, N/m

T_r = -\frac{1}{2} \times 252 \, N/m \times (0.118 \, m)^2 = -1.756 \, J

(c) \, T_t = T_g + T_r

= 0.304 \, J + (-1.756 \, J) = -1.452 \, J

E_k = \frac{1}{2} \times m \times v^2

\frac{1}{2} \times 0.263 \, kg \times v^2 = 1.452 \, J

v = \sqrt{\frac{2 \times 1.452 \, J}{0.263 \, kg}} = 3.67 \, m/s

(d) \, \text{Si se duplica la velocidad inicial, la energía cinética se cuadruplica:}

E_k = 4 \times 2.06 \, J = 8.24 \, J

E_k = \frac{1}{2} \times k \times x^2

x = \sqrt{\frac{2 \times 8.24 \, J}{252 \, N/m}} = 0.235 \, m = 23.5 \, cm

\text{Respuesta: }( a ) \, 0.304 \, J, (b) \, -1.756 \, J, (c) \, 3.67 \, m/s, (d) \, 23.5 \, cm

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