A toy gun uses a spring to project a 4.5-g soft rubber sphere horizontally. The spring constant is 8.0 N/m, the barrel of the gun is 13 cm long, and a constant frictional force of 0.035 N exists between barrel and projectile. With what speed does the projectile leave the barrel if the spring was compressed 5.8 cm for this launch? (Assume the projectile is in contact with the barrel for the full 13 cm.)

1.93 m/s

Explanation:

Parameters given:

Mass = 4.5g = 0.0045kg

Spring constant = 8.0 N/m

Length of barrel = 13 cm = 0.013m

Frictional force = 0.035N

Compression = 5.8 cm = 0.058m

First, we find the P. E. stored in the spring:

P. E. = ½*k*x²

P. E. = ½ * 8 * 0.058² = 0.013J

Then, we find the work done by the frictional force while the sphere is leaving the barrel of the gun:

Work = Force * distance

The distance here is the length of the barrel.

Work = 0.035 * 0.13 = 0.0046 J

The kinetic energy of the sphere can now be found:

K. E. = P. E. - Work done

K. E. = 0.013 - 0.0046 = 0.0084J

We can now find the speed using the formula for K. E.:

K. E. = ½*m*v²

0.0084 = ½ * 0.0045 * v²

v² = 0.0084/0.00255 = 3.733

=> v = 1.93 m/s

Answer: v = 4.97 m/s

Explanation: mass of spring (m) = 4.5g = 4.5/1000 = 0.0045 kg

Spring constant (k) = 8.0 N/m

Length of barrel = distance traveled when projectile was launched = 13cm = 0.13 m

Frictional force = 0.035 N

Extension (x) = 5.8cm = 0.058 m

Assuming the motion is of a constant acceleration, newton's laws of motion is valid

F - fr = ma

Where F = applied force that propelled the projectile = kx = 8*0.058 = 0.464 N

a = acceleration of object.

By substituting the parameters, we have

0.464 - 0.035 = 0.0045 a

0.429 = 0.0045 a

a = 0.429/0.0045

a = 95.3 m/s²

The bullet started from rest, hence initial velocity (u) equals zero.

To get the final velocity v, we use

v² = u² + 2aS, where S = length of barrel or distance covered by projectile.

v ² = 0² + 2 (95.3) (0.13)

v² = 2 (95.3) (0.13)

v² = 24.778

v = √24.778

v = 4.97 m/s