# A surfer is catching a wave. Suppose she starts at the top of the wave with a speed of 1.3 m/s...

## Question:

A surfer is catching a wave. Suppose she starts at the top of the wave with a speed of 1.3 m/s and moves down the wave until her speed increases to 8.0 m/s. The drop in her vertical height is 2.8 m. If her mass is 59 kg, how much work is done by the (nonconservative) force of the wave?

## Energy Conservation:

The energy of the system remains conserved at the different instant of the motion. It only changes its form from one to another. As the height of the object decreases, the potential energy decreases and the kinetic energy will increase.

Given Data:

• Initial speed of the surfer {eq}\rm (u) = 1.3 \ m/s {/eq}
• Final speed of the surfer {eq}\rm (v) = 8 \ m/s {/eq}
• Change in the height {eq}\rm (h_{1} - h_{2}) = 2.8 \ m {/eq}

The initial energy of the surfer

{eq}\rm E_{1} = \dfrac{1}{2}mu^{2} + mgh_{1} {/eq}

Now, the final energy of the surfer {eq}\rm E_{2} = \dfrac{1}{2}mv^{2} + mgh_{2} \\ {/eq}

Now the work done by the Conservative force

{eq}\rm W = E_{2} - E_{1} \\ W = \dfrac{1}{2}mv^{2} + mgh_{2} - \dfrac{1}{2}mv^{2} - mgh_{1} \\ W = \dfrac{1}{2}m(u^{2} -v^{2}) + mg(h_{1} - h_{2}) \\ W = \dfrac{1}{2} \times 59 (8^{2} - 1.3^{2}) + 59 \times 9.8 \times (-2.8 \ m ) \\ W = 219.185\ J {/eq} 