If it takes an average force of 5.4 N to push a 4.8-g dart 5.4 cm into a dart gun, assuming the barrel is frictionless, how fast will the dart exit the gun?

F•s= m•v²/2

v = sqrt(2•F•s/m) =
=sqrt(2•5.4•5.4•10^-2/4.8•10^-3)= 11 m/s

To find the speed at which the dart will exit the gun, we can use the principle of conservation of energy. The work done in pushing the dart into the gun is equal to the kinetic energy of the dart as it exits the gun.

The work done is given by the formula:

Work = Force * Distance

Here, the force is 5.4 N, and the distance the dart is pushed into the gun is 5.4 cm. However, we need to convert the distance to meters before calculating the work. Therefore, the distance becomes 0.054 m.

Work = 5.4 N * 0.054 m = 0.2916 J (Joules)

This work done is equal to the kinetic energy of the dart as it exits the gun. The kinetic energy of an object is given by the formula:

Kinetic Energy = (1/2) * mass * velocity^2

We already know the mass of the dart, which is 4.8 g. But we need to convert it to kilograms, so the mass becomes 0.0048 kg.

Substituting the given values into the formula, we have:

0.2916 J = (1/2) * 0.0048 kg * velocity^2

Simplifying the equation, we have:

0.2916 J = 0.0024 kg * velocity^2

Dividing both sides of the equation by 0.0024 kg:

velocity^2 = (0.2916 J) / (0.0024 kg)
velocity^2 = 121 J/kg

Taking the square root of both sides of the equation to solve for velocity:

velocity = √(121 J/kg)
velocity = 11 m/s

Therefore, the dart will exit the gun with a speed of 11 m/s.