You are designing a delivery ramp for crates containing exercise equipment. The 1470-N crates will move at 1.8 m>s at the top of a ramp that slopes downward at 22.0°. The ramp exerts a 515-N kinetic friction force on each crate, and the maximum static friction force also has this value. Each crate will compress a spring at the bottom of the ramp and will come to rest after traveling a total distance of 5.0 m along the ramp. Once stopped, a crate must not rebound back up the ramp. Calculate the largest force constant of the spring that will be needed to meet the design criteria.

The given data can be listed below,

• The velocity of the crate is,$v=1.8\text{\hspace{0.17em}m/s}$
• The angle of inclination is,$\theta ={22}^{\circ }$
• The friction force on the ramp is,$f_k=515\text{\hspace{0.17em}N}$
• The distance travelled by the crate along with ramp is,$d=5\text{\hspace{0.17em}m}$

When one item is exerting driving force over another, kinetic friction occurs at the point of contact between the two objects.

At the bottom of the ramp, where the block will be halted and moving at a stop, the kinetic energy and gravitational potential energy are both zero. Additionally, the zero level to be there, so the gravitational potential energy is also zero at that location. The kinetic friction force now exerts its force in the direction that is counter to the displacement.

The energy of the system is given by,

$\begin{array}{c}{E}_f-E_i=F_{Other}\\ K_f+U_{grav,f}+U_s-(K_i+U_{grav,i})=f_{k}s\end{array}$

Substitute all the values in the above,

$\begin{array}{c}0+0+\frac{1}{2}k{x}^2-(\frac{1}{2}m{v}_1^2+mgh)=(-515\text{\hspace{0.17em}N})\left(5\text{\hspace{0.17em}m}\right)\\ =-2575\text{\hspace{0.17em}J}\\ \frac{1}{2}k{x}^2-\frac{1}{2}\frac{1470\text{\hspace{0.17em}N}}{9.8{\text{\hspace{0.17em}m/s}}^2}{\left(1.8\text{\hspace{0.17em}m/s}\right)}^2-\frac{1470\text{\hspace{0.17em}N}}{g}g\left(5\sin 22\right)=-2575\text{\hspace{0.17em}J}\\ \frac{1}{2}k{x}^2-2996.3\text{\hspace{0.17em}J}=-2575\text{\hspace{0.17em}J}\\ k{x}^2=842\text{\hspace{0.17em}J}\end{array}$

The force on the spring is given by,

$f_s=f_{static,\max }$

Substitute all the values in the above,

$kx=515\text{\hspace{0.17em}N}$

So, the stretch is given by,

$\begin{array}{c}\left(kx\right)x=842\text{\hspace{0.17em}J}\\ \text{x}=\frac{842\text{\hspace{0.17em}J}}{515\text{\hspace{0.17em}N}}\\ =1.635\text{\hspace{0.17em}m}\end{array}$

The force constant is given by,

$\begin{array}{c}k\left(1.635\text{\hspace{0.17em}m}\right)=515\text{\hspace{0.17em}N}\\ k=\frac{515\text{\hspace{0.17em}N}}{1.635\text{\hspace{0.17em}m}}\\ =315\text{\hspace{0.17em}N/m}\end{array}$

Thus, the maximum force constant of the spring $315\text{\hspace{0.17em}N/m}$