Calculating a force constant using kinetic and potential energy?

[erik-the-red]

You are designing a delivery ramp for crates containing exercise equipment. The crates weighing $$1500 N$$ will move at a speed of $$2.00 m/s$$ at the top of a ramp that slopes downward at an angle $$24.0^\circ.$$ The ramp exerts a kinetic friction force of $$540 N$$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 $$7.90 m$$ along the ramp. Once stopped, a crate must not rebound back up the ramp.

Question:

Calculate the force constant of the spring that will be needed in order to meet the design criteria.

My answer is 27.4 N/m, but I am not quite sure that it is correct.

I know $$K_1=(1/2)(1500/9.80)(2.00^2)=306 J$$

$$K_2$$ is zero. $$U_1$$ is also zero. $$W_f$$ is $$-4270 J$$. I define $$x_2=7.90m$$.

So, I should use the equation $$K_1+U_1+W_f=mgy_2+(1/2)(k)(x_2)^2$$.

$$y_2=x_2*sin(24.0^\circ)=3.21$$. But, it is -3.21 because $$y_2$$ is below $$y_1=0$$.

Using the equation I believe I should use, I got the answer to be $$k=27.4 m/s.$$

 

[anathema]

Your calculations and approach seem correct. The only thing I would suggest is to double check your unit conversions. You have used meters for distance and seconds for time, but the units for force constant (k) should be in N/m. Make sure all your units are consistent throughout your calculations. Other than that, your answer of 27.4 N/m seems reasonable.

 

[quicknote]

Your calculations and approach seem to be correct. The force constant of the spring can be calculated using the equation K_1 + U_1 + W_f = mgy_2 + (1/2)k(x_2)^2, where K_1 is the initial kinetic energy, U_1 is the initial potential energy, W_f is the work done by friction, m is the mass of the crate, g is the acceleration due to gravity, y_2 is the final height of the crate, and x_2 is the distance traveled along the ramp.

Plugging in the given values, we get:

306 J + 0 J + (-4270 J) = (1500 N)(9.8 m/s^2)(-3.21 m) + (1/2)k(7.90 m)^2

Solving for k, we get k = 27.4 N/m, which is the same as your answer. This means that in order for the crates to come to a complete stop without rebounding, the spring must have a force constant of 27.4 N/m.