Question

A perfect spring whose spring constant is 180 newtons per meteris attached to a 3.2-kilogram cart that slides on a frictionlesshorizontal surface, as shown. The cart is moved 40 centimetersfrom its equilibrium position and released. Find the speed of thecar when it passes its equilibrium position.

Answer 1

Given data:

The value of the spring constant is,

`k=180Nm^(-1)`

The value of the mass of the cart is,

`m=3.2\text{ kg}`

The distance moved by the cart from the equilibrium is,

`\begin{gathered} d=40\text{ cm} \\ d=0.4\text{ m} \end{gathered}`

The acceleration acquired by the cart is,

`\begin{gathered} kd=ma \\ 180*0.4=3.2* a \\ 72=3.2* a \\ a=22.5ms^(-1) \end{gathered}`

The initial velocity of the cart is,

`u=0ms^(-1)`

The velocity of the cart when it passes the equilibrium is,

`\begin{gathered} v=u+2ad \\ v=0+2*22.5*0.4 \\ v=18ms^(-1) \end{gathered}`

Thus, the velocity of the cart is 18 meter per second.