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.

Question

12-12-2019
Physics

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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.

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davidlcastiblanco davidlcastiblanco · Answer 1 · 2019-12-17T01:50:47+01:00

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

okpalawalter8 okpalawalter8 · Answer 2 · 2022-03-24T12:43:28+01:00

The largest force constant of the spring is mathematically given as

K = 25351. 69 N / m

What is the largest force constant of the spring that will be needed to meet the design criteria?

Question Parameters:

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

traveling a total distance of 5.0 m

Generally, the equation for the work done by all forces is mathematically given as

Ws + We - xk = Em - Ex

Therefore

0.5 * K * x^2 = 150*9.8* 5* sin (22°) - 5150* 5 + 0.5*150*(1.8m/s)^2

K * x^2 = 421.358

Hence

K * x = 515 + 150*9.8* sin(22°)

K * x = 3268.35

In conclusion, Comparing both equations

K * x^2 = 421.358

K * x = 3268.35

x = 0.1289

K = 25351. 69 N / m