Question

You are designing a delivery ramp for crates containing exercise equipment. The 1890 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.

Answer 1

Answer:

K = 588.3 N/m

Explanation:

From a forces diagram, and knowing that for the maximum value of K, the crate will try to rebound back up (Friction force will point downward):

Fe - Ff - W*sin(22) = 0    Replacing Fe = K*X   and then solving for X:

$X = \frac{Ff + W*sin(22)}{K}=\frac{1223}{K}$

By conservation of energy:

$\frac{K*X^{2}}{2}-mg*d*sin(22)-\frac{m*V^{2}}{2}=-Ff*d$

Replacing our previous value for X and solving the equation for K, we get maximum value to prevent the crate from rebound:

K = 588.3 N/m