Question

The spring in FIGURE CP10.73 has a spring constant of $1000N/m$. It is compressed $15cm$, then launches a $200g$block. The horizontal surface is frictionless, but the block’s coefficient of kinetic friction on the incline is $0.20$. What distance d does the block sail through the air?

Short answer

The block sails in the air to a distance of$6.676m$

Step-by-step solution

Given information

Spring constant $=1000N/m$

Compression of spring$=15cm$

The weight of the block$=200g$

Coefficient of kinetic friction$=0.20$

Explanation

According to the conservation of energy, the energy equation is

$$\begin{gathered} K_i+U_{gi}+W_{ext}=K_f+U_{gf}+E_f \\ \frac{1}{2}k{\left(x_i-x_f\right)}^2=\frac{1}{2}m{v}_f^2+mg{y}_f+f_k\Delta x \\ \text{where}{f}_k\text{isfrictionalforceand}\Delta x\text{isthedistancealongtheslope} \\ {f}_k={\mu }_{k}n \\ \text{Fromthegivenfigurenormalforce}n=mg\cos \left(45°\right)\text{and}\Delta x=\frac{\mathrm{y}}{\sin \left(45°\right)} \end{gathered}$$

$$\begin{gathered} \frac{1}{2}k{\left(x_i-x_f\right)}^2=\frac{1}{2}m{v}_f^2+mg{y}_f+f_k\Delta x \\ \frac{1}{2}m{v}_f^2=\frac{1}{2}k{\left(x_i-x_f\right)}^2-mg{y}_f-f_k\Delta x \\ {v}_f=\sqrt{\left(\frac{2}{m}\right)\left(\frac{1}{2}k{\left(x_i-x_f\right)}^2-mg{y}_f-\frac{m{\mu }_{k}g{y}_f\cos 45°}{\sin 45°}\right)} \\ {v}_f=\sqrt{\left(\frac{k}{m}{\left(x_i-x_f\right)}^2-2g{y}_f-\frac{2{\mu }_{k}g{y}_f\cos 45°}{\sin 45°}\right)} \\ {v}_f=\sqrt{\left(\left[\frac{\left(1000N/m\right){\left(0.15m\right)}^2}{0.2kg}\right]-\left[\left(2\right)\left(9.8m/s^2\right)\left(2m\right)\right]-\left[2\left(0.2m\right)\left(9.8m/s^2\right)\left(2m\right)cot(45°)\right]\right)} \\ =8.091m/s \end{gathered}$$

By using the equation of motion:

$$\begin{gathered} y=v_{y}t-\frac{1}{2}g{t}^2 \\ Theverticaldis\tan cey=0 \\ {v}_y=v_f\sin (45°) \\ 0=\left(v_f\sin (45°)\times t\right)-\frac{1}{2}g{t}^2 \\ \frac{1}{2}g{t}^2=v_f\sin \left(45°\right)\times t \\ t=\frac{2v_f\sin \left(45°\right)}{g} \\ t=\frac{2\left(8.091m/s\right)}{\left(9.8m/s^2\right)}\left(\frac{1}{\sqrt{2}}\right) \\ t=1.167s \\ \text{Horizontal distance,}x=v_x\times t\text{.} \\ {v}_x=v_f\cos \left(45°\right) \\ d=v_f\cos \left(45°\right)\times t \\ =\left(8.091\mathrm{m}/\mathrm{s}\right)\left(\frac{1}{\sqrt{2}}\right)\left(1.167\mathrm{s}\right) \\ =6.676\mathrm{m} \end{gathered}$$