Question

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.

Answer 1

Answer:

K = 25351. 69 N / m

Explanation:

Given : Fk = 515 N , v = 1.8 m /s  , d = 5.0 m , β = 22.0 ° , m = 150 kg

Using the work done by all forces at initial and the end can determine the constant of the spring

Ws + We - Fk = Em - Ef

- ¹/₂ * K * x²  + m*g*h - F*d = 0 - ¹/₂ * m * v²

Also the round motion part

K* x = F + We

K * x = F + m*g*h

Replacing numeric to equal the equations and find the constant

¹/₂ * K * x² = 150*9.8* 5* sin (22°) - 5150* 5 + ¹/₂*150*(1.8m/s)²

K * x² = 421.358

Now use the other equation

K * x = 515 + 150*9.8* sin(22°)

K * x = 3268.35

Both equation give x' as a

x = 0.1289 m now using in any equation can find K

K = 25351. 69 N / m