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Title: What is the velocity of the block before it hits the spring? A 210 g block is dropped onto a relaxed vertical? Post by: rk_9879 on Nov 2, 2012 A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.5 N/cm. The block becomes attached to the spring and compresses the spring 12 cm before momentarily stopping.
What is the velocity of the block before it hits the spring? Title: What is the velocity of the block before it hits the spring? A 210 g block is dropped onto a relaxed vertical? Post by: Smoren on Nov 2, 2012 This is an energy question; the energy of the block is kinetic energy and the energy for the spring is compressed energy. In this problem, when the spring is compressed 12 cm all of the kinetic energy of the block has been used to compress the spring and the blocks kinetic energy = 0. So the energy of the spring at 12 cm is equal to the energy of the block just before it started to compress the spring or:
KE(b) = 0.5 * k * x^2; KE(b) = energy of the block just before it started to compress the spring, k = spring constant = 2.5N/cm, and x = total compression of the spring = 12cm. Also note that KE(b) = 0.5 * M * V(b)^2; M = 0.210kg. This gives: 0.5 * M * V(b)^2 = 0.5 * k * x^2; solve for V(b) = velocity of the block just before it contacts the spring V(b)^2 = ( k * x^2) / M V(b) = sqrt[( k * x^2) / M]; plugging in the numbers gives: V(b) = sqrt[(2.5kg*m/(s^2 * cm) * (12cm)^2 / 0.210kg] V(b) = sqrt[1714.29m*cm / s^2]; note that prior to completing the sqrt you need to convert cm to m using the fact that 1 = 1m / 100cm. This gives: V(b) = sqrt[(714.29m*cm / s^2) * (1m / 100cm) V(b) = sqrt(17.1429m^2 / s^2) V(b) = 4.140 m/s = the velocity of the block just before it hits the spring. Note that one of the key issues on this problem was converting units so they would be compatible in the final answer. This is one of the reasons it is always important to always include units in any calculation. I hope this helps. |