The spring is released and a 0.10-kilogram plastic sphere is fired from the launcher. Calculate the maximum speed with which the plastic sphere will be launched. [Neglect friction.] [Show all work, including the equation and substitution with units.]

A toy launcher that is used to launch small plastic spheres horizontally contains a spring with a spring constant of 50. newtons per meter. The spring is compressed a distance of 0.10 meter when the launcher is ready to launch a plastic sphere.

A) Determine the elastic potential energy stored in the spring when the launcher is ready to launch a plastic sphere.

B) The spring is released and a 0.10-kilogram plastic sphere is fired from the launcher. Calculate the maximum speed with which the plastic sphere will be launched. [Neglect friction.] [Show all work, including the equation and substitution with units.]

A) The elastic potential energy (EPE) is calculated as follows:

EPE = 1/2 · k · x²

Where:

k = spring constant.

x = compressing distance

EPE = 1/2 · 50 N/m · (0.10 m)²

EPE = 0.25 J

The elastic potential energy stored in the spring when it is compressed 0.10 m is 0.25 J.

B) Since there is no friction, all the stored potential energy will be converted into kinetic energy when the spring is released. The equation of kinetic energy (KE) is the following:

KE = 1/2 · m · v²

Where:

m = mass of the sphere.

v = velocity

The kinetic energy of the sphere will be equal to the initial elastic potential energy:

KE = EPE = 1/2 · m · v²

0.25 J = 1/2 · 0.10 kg · v²

2 · 0.25 J / 0.10 kg = v²

v = 2.2 m/s

The maximum speed of the plastic sphere will be 2.2 m/s

andromache andromache · Answer 2 · 2022-02-09T07:02:57+01:00

A) The elastic potential energy stored in the spring when it is compressed 0.10 m is 0.25 J.

B) The maximum speed of the plastic sphere will be 2.2 m/s

Computation of the elastic potential energy & maximum speed. :

A) The elastic potential energy should be

[tex]EPE = 1\div 2 \times k \times x^2[/tex]

Here

k = spring constant.

x = compressing distance

So,

[tex]EPE = 1\div 2 \times 50 N/m \times (0.10 m)^2[/tex]

EPE = 0.25 J

b. Now the maximum speed should be

[tex]KE = 1\div 2 \times m \times v^2[/tex]

Here

m = mass of the sphere.

v = velocity

So,

[tex]0.25 J = 1/2 \times 0.10 kg \times v^2\\\\2 \times 0.25 J \div 0.10 kg = v^2[/tex]

v = 2.2 m/s

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