Answer to Question #144934 in Physics for Dunia

Question

105. A 1.3-kg block is pushed up against a stationary spring compressing it a distance of 4.2 cm. When the block is released, the spring pushes it away across a frictionless, horizontal surface. What is the speed of the block, given that the spring constant of the spring is 1400 N/m?

Answer 1

After the block is pushed up against a stationary spring, the potential energy accumulated in the spring is:

$$U = \dfrac{kx^2}{2}$$

where $k = 1400N/m$ is the spring constant and $x = 4.2cm = 0.042m$ is the distance of compressing. After the block was released, all this energy was converted to the kinetic energy of the block:

$$T = \dfrac{mv^2}{2} = U$$

where $m = 1.3kg$ is the mass of the block. Substituting the expression for $U$, obtian:

$$\dfrac{mv^2}{2} = \dfrac{kx^2}{2}\\ \space \\ v = x\sqrt{\dfrac{k}{m}}\\ v = 0.042\sqrt{\dfrac{1400}{1.3}} \approx 1.38m/s$$

Answer. 1.38 m/s.