Question

A block of mass, m, is pushed up against a spring, compressing it a distance x, and is then released. The spring projects the block along a frictionless horizontal surface, giving the block a speed v. The same spring projects a second block of mass 4m, giving it a speed 3v. What distance was the spring compressed in the second case?

Answer 1

Answer:

x₂ =6*x₁

Explanation:

• In absence of friction, along a horizontal surface, all the elastic potential energy stored in the spring becomes kinetic energy, so we can write the following expression for the first case:

       $\frac{1}{2}*k*x_{1}^{2} = \frac{1}{2}*m*v^{2} (1)$

• In the second case, we have a mass 4m, which reaches to a speed 3v due to the energy stored in the spring.
• We can write the same equation than for (1) replacing m by 4m and v by 3v, as follows:

        $\frac{1}{2}*k*x_{2} ^{2} = \frac{1}{2} * 4*m*(3*v)^{2} =\frac{1}{2}*4*m*9*v^{2} = \frac{1}{2}* 36 m*v^{2} (2)$

• Dividing (2) over (1) on both sides, and rearranging terms, we get:

        $x_{2} ^{2} = 36 * x_{1} ^{2}$

• Taking square roots on both sides, we get:
• x₂ = 6*x₁
• So, in this case, the spring was compressed six times the distance that was compressed in the first case.