Question

A person exerts a 19-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 Part A Determine the spring constant of the spring. Express your answer to two significant figures and include the appropriate units. k- 320 듦 Submit My Answers Give Up Correct Part B Determine the total energy of the system. Express your answer to two significant figures and include the appropriate units Utotal 0.57 J Submit My Answers Give Up Correct Part C Write an expression (t) for the motion of the cart. Assume the frequency1 Hz Express your answer in terms of the variable t and constant π z(t)

Answer 1

To solve this problem, apply the concepts related to Hooke's law. From there we will find the spring constant. Subsequently, applying Energy balance, which includes gravitational potential energy, elastic potential energy and kinetic energy, we will bury the system's energy. Finally, using the displacement expression for the simple harmonic movement, we will find the expression that describes the system.

PART A) The expression for the spring force is

$F=kx$

Here,

k = Spring constant

x = Displacement

Rearranging to find the spring constant we have that

$k = \frac{F}{x}$

$k = \frac{19}{0.06}$

$k = 316.66N/m \approx 320N/m$

PART B ) The gravitational potential energy acts on the spring holds the cart is zero. Since cart is placed in the equilibrium position. The kinetic energy of the cart is zero.  Therefore the expression for the total energy is,

$E = (PE)_g+(PE)_{spring}+KE$

$E = 0+\frac{1}{2} kx^2+0$

$E = \frac{1}{2} (316.66N/m)(0.06)^2$

$E = 0.569J \approx 0.57J$

PART C) The expression for the angular frequency is

$\omega = 2\pi f$

$\omega = 2\pi (1Hz)$

$\omega = 2\pi rad/s$

The equation for the motion of the cart is

$x(t)= Acos(\omega t)$

Replacing,

$x(t) = 0.06 cos (2\pi t)$