Question

A person exerts a 14-N force on a cart attached to a spring and holds the cart steady. The cart is displaced 0.060 m from its equilibrium position. When the person stops holding the cart, the system cart spring undergoes simple harmonic motion.
1. Determine the spring constant of the spring.2. Determine the total energy of the system.3. Write an expression x(t) for the motion of the cart. Assume the frequency f=1Hz. Express your answer in terms of the variable t and constant pi.

Answer 1

Answer:

1.  $k=233.33\ N.m^{-1}$

2. $PE_s=0.42\ J$

3. $x(t)=0.06\ \cos\ (2\pi.t)$ is the expression for the motion of the cart.

Explanation:

Given:

force applied to hold the cart attached to the spring, $F=14\ N$

displacement of the cart under the force, $\Delta x=0.06\ m$

1.

Now we know that the spring constant is given as:

$F=k.\Delta x$

$14=k\times 0.06$

$k=233.33\ N.m^{-1}$

2.

The total energy can be given as the maximum spring potential because this the energy introduced initially to the spring which takes other form of energy be it heat or kinetic energy.

The spring potential is given as:

$PE_s=\frac{1}{2} .k.\Delta x^2$

$PE_s=0.5\times 233.33\times 0.06^2$

$PE_s=0.42\ J$

3.

We know the genral equation of the spring oscillation as:

$x(t)=A\ \cos\ (\omega.t)$ ......................(1)

Given that the frequency, $f=1\ Hz$

The expression for the frequency is given as:

$\omega=2\pi f$

$\omega=2\pi\times 1$

$\omega=2\pi$

Now from eq. (1)

$x(t)=0.06\ \cos\ (2\pi.t)$ is the expression for the motion of the cart.