Question

a block of mass ( m = 0.2 \text{kg} ) on a ramp with negligible friction is pressed against a spring with spring constant ( k = 800 \text{n/m} ), compressing it a distance ( delta x = 0.1 \text{m} ). the block is then released and travels up the ramp. what is the height of the block above its initial height when it has speed ( v = 5.0 \text{m/s} )?

a ( 0.75 \text{m} )
b ( 1.5 \text{m} )
c ( 2.0 \text{m} )
d ( 3.25 \text{m} )

Explanation:

Step1: Apply conservation of mechanical energy

The initial elastic potential energy of the spring is converted into kinetic energy and gravitational potential energy of the block. The formula for conservation of energy is \(E_{initial}=E_{final}\). The initial energy is the elastic potential energy of the spring \(E_{elastic}=\frac{1}{2}k(\Delta x)^{2}\), and the final energy is the sum of kinetic energy \(E_{k}=\frac{1}{2}mv^{2}\) and gravitational potential energy \(E_{g}=mgh\). So, \(\frac{1}{2}k(\Delta x)^{2}=\frac{1}{2}mv^{2}+mgh\).

Step2: Solve for \(h\)

First, substitute the given values \(m = 0.2\space kg\), \(k = 800\space N/m\), \(\Delta x=0.1\space m\), \(v = 5.0\space m/s\) into the equation \(\frac{1}{2}k(\Delta x)^{2}=\frac{1}{2}mv^{2}+mgh\).

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Answer:

A. \(0.75\space m\)