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andreyandreev [35.5K]
3 years ago
6

A spring driven dart gun propels a 10g dart. It is cocked by exerting a force of 20N over a distance of 5cm. With what speed wil

l the dart leave the gun, assuming the spring has negligible mass?

Physics
2 answers:
adelina 88 [10]3 years ago
6 0
<span>14 m/s Assuming that all of the energy stored in the spring is transferred to dart, we have 2 equations to take into consideration. 1. How much energy is stored in the spring? 2. How fast will the dart travel with that amount of energy. As for the energy stored, that's a simple matter of multiplication. So: 20 N * 0.05 m = 1 Nm = 1 J For the second part, the energy of a moving object is expressed as KE = 0.5 mv^2 where KE = Kinetic energy m = mass v = velocity Since we now know the energy (in Joules) and mass of the dart, we can substitute the known values and solve for v. So KE = 0.5 mv^2 1 J = 0.5 0.010 kg * v^2 1 kg*m^2/s^2 = 0.005 kg * v^2 200 m^2/s^2 = v^2 14.14213562 m/s = v So the dart will have a velocity of 14 m/s after rounding to 2 significant figures.</span>
kolezko [41]3 years ago
5 0

The dart will leave the gun with speed 10 m/s

\texttt{ }

<h3>Further explanation</h3>

<em>Hooke's Law states that the length of a spring is directly proportional to the force acting on the spring.</em>

\boxed {F = k \times \Delta x}

<em>F = Force ( N )</em>

<em>k = Spring Constant ( N/m )</em>

<em>Δx = Extension ( m )</em>

\texttt{ }

The formula for finding Young's Modulus is as follows:

\boxed {E = \frac{F / A}{\Delta x / x_o}}

<em>E = Young's Modulus ( N/m² )</em>

<em>F = Force ( N )</em>

<em>A = Cross-Sectional Area ( m² )</em>

<em>Δx = Extension ( m )</em>

<em>x = Initial Length ( m )</em>

Let us now tackle the problem !

\texttt{ }

<u>Given:</u>

compression of the spring = x = 5 cm = 0.05 m

magnitude of the force = F = 20 N

mass of dart = m = 10 g = 0.01 kg

<u>Asked:</u>

speed of the dart = v = ?

<u>Solution:</u>

<em>We will use </em><em>Conservation of Energy </em><em>formula to solve this problem as follows:</em>

Ep = Ek

\frac{1}{2} F x = \frac{1}{2} m v^2

F x = m v^2

v^2 = F x \div m

v = \sqrt { F x \div m }

v = \sqrt { 20 \times 0.05 \div 0.01 }

v = \sqrt { 100 }

v = 10 \texttt{ m/s}

\texttt{ }

<h3>Learn more</h3>
  • Young's modulus : brainly.com/question/6864866
  • Young's modulus for aluminum : brainly.com/question/7282579
  • Young's modulus of wire : brainly.com/question/9755626

\texttt{ }

<h3>Answer details</h3>

Grade: College

Subject: Physics

Chapter: Elasticity

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