Chapter 10: Q. 60 (page 259)

A clever engineer designs a “sprong” that obey the force law $F_{x} = - q {(x - x_{e q})}^{3}$ ,where $x_{e q}$ is the equilibrium position of the end of the sprong and q is the sprong constant. For simplicity, we’ll let $x_{e q} = 0 m$ . Then $F_{x} = - q x^{3}$ .
a. What are the units of q?
b. Find an expression for the potential energy of a stretched or compressed sprong.
c. A sprong-loaded toy gun shoots a $20 g$ plastic ball. What is the launch speed if the sprong constant is $40, 000$ , with the units you found in part a, and the sprong is compressed $10 c m$ ? Assume the barrel is frictionless.

Short Answer

Expert verified

$(a)Unitofqis N / m^{3} (b)Potentialenergy, U (x) = \frac{q x^{4}}{4} (c)Launchspeedis 10 m / s .$

Step by step solution

Given information (part a)

The "Sprong" obey the force law $F_{x} = - q {(x - x_{e q})}^{3}$

Where;

$q =$ the sprong constant.

$x_{e q} =$ the equilibrium position.

$x_{e q} = 0$

Thus; $F_{x} = - q x^{3}$

Mass of plastic ball $= 20 g$

The sprong constant $= 40, 000 .$

The compression of sprong $= 10 c m$

Explanation (part a)

For the unit of q,

Consider the equation,

$F_{x} = - q x^{3}$

$F_{x}$ is in Newton (N)

x is in meters (m)

Hence;

$F_{x} = - q x^{3} N = q {(m)}^{3}$

for the unit, the sign does not matter

localid="1649412728684" $q = \frac{N}{m^{3}}$

Given information (part b)

The "Sprong" obey the force law $F_{x} = - q {(x - x_{e q})}^{3}$

Where;

$q =$ the sprong constant

$x_{e q} =$ the equilibrium position.

$x_{e q} = 0$

Thus; $F_{x} = - q x^{3}$

Mass of plastic ball $= 20 g$

The sprong constant $= 40, 000 .$

The compression of sprong $= 10 c m$

Explanation (part b)

The potential energy of a sprong:

U (x) = - \int F_{x} d x = - \int - q x^{3} d x = \int q x^{3} d x = q \int x^{3} d x = q [\frac{x^{4}}{4}] U (x) = \frac{q x^{4}}{4}

Given information (part c)

"Sprong" obey the force law $F_{x} = - q {(x - x_{e q})}^{3}$

Where;

$q =$ the sprong constant

$x_{e q} =$ the equilibrium position.

$x_{e q} = 0$

Thus; $F_{x} = - q x^{3}$

Mass of plastic ball $= 20 g$

The sprong constant $= 40, 000 .$

The compression of sprong $= 10 c m$

Explanation (part c)

Applying the energy conservation equation to the ball and sprong system:

$K_{i} + U_{i} = K_{f} + U_{f} \frac{1}{2} m v_{f}^{2} + (0 J) = (0 J) + \frac{q x_{i}^{4}}{4} \Rightarrow v_{f} = \sqrt{\frac{q x_{i}^{4}}{2 m}} v_{f} = \sqrt{\frac{q x_{i}^{4}}{2 m}} = \sqrt{\frac{(40, 000 N / m^{3}) (10 cm)^{4}}{(2) (20 g)}} = \sqrt{\frac{(40, 000 N / m^{3}) \times {(0.1 m)}^{4}}{2 \times (\frac{20}{1000} Kg)}} = 10 m / s$