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

The distance the spring was compressed in the second case, where the mass was four times the original mass and the speed was three times the original speed, would be three times the compression distance of the first case.

Step-by-step explanation:

The question is asking to determine the distance a spring was compressed to launch a second block with a mass four times larger than the first block and at a speed three times greater than the first block. We can use the conservation of energy principle, specifically the relationship between the potential energy stored in the spring (spring potential energy) and the kinetic energy of the block.

The potential energy in a compressed spring is given by ½ kx², where k is the spring constant and x is the compression distance. The kinetic energy of a block is given by ½ mv², where m is the mass and v is the velocity of the block

For the first block, if we assume the spring's potential energy is fully converted to the block's kinetic energy, the equation becomes ½ kx² = ½ mv². And for the second block of mass 4m launched at 3v, the equation is ½ kx'² = ½ –4m(3v)². By setting up a ratio of the spring potential energies, which must be equal to the ratio of the kinetic energies as the system is frictionless, we find that the compression distance x' for the second block is equal to 3x.

Therefore, if the spring was originally compressed by a distance x to launch the first block, it would have been compressed by 3x to launch the second block that has four times the mass and is given three times the velocity.

Answer 2

x₂ =6*x₁

Step-by-step 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:

`(1)/(2)*k*x_(1)^(2) = (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:

`(1)/(2)*k*x_(2) ^(2) = (1)/(2) * 4*m*(3*v)^(2) =(1)/(2)*4*m*9*v^(2) = (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.