Problem 7.48 You are designing a delivery ramp for crates containing exercise eq

Question

Problem 7.48 You are designing a delivery ramp for crates containing exercise equipment. The 1670-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. Part A Calculate the largest force constant of the spring that will be needed to meet the design criteria. Express your answer with the appropriate units. k= 1 Value Units Submit Give Up

Explanation / Answer

The work/energy theorem says: The change in an object's KE equals the work done on the object by all forces.

Change in crate's KE = ½mv² = ½(W/g)v² =-275.7m/s

Work done by gravity = Weight × drop in height = W(Lsin) =3127.96J

Work done by friction = Ff× L (negative because it's opposite from the direction of motion) =-2575N

Work done by spring = ½kx² (negative because it's opposite from the direction of motion).
Put them all together:
½(W/g)v² = W(Lsin) (Ff)L ½kx²

But this equation still has two unknowns ("k" and "x"), so this is still not enough to solve uniquely for "k". So we must make use of this additional information:
Basically that means that, once the spring has compressed to distance "x", the combined (downward) forces of gravity and static friction (Fstat) must be enough to match the (upward) push of the spring. That is, we must have:

Wsin + Fstat = kx
or:
Fstat = kx Wsin

The MAXIMUM value of Fstat is given as 515 N

This is greater than or equal to the ACTUAL static friction:
Fstat_max Fstat

So substituting:
Fstat_max kx Wsin

And rearranging:
kx Fstat_max + Wsin

Now recall that equation we derived before:
½(W/g)v² = W(Lsin) (Ff)L ½kx²
Rewrite that as:
½kx² = WLsin (Ff)L + ½(W/g)v²

Multiply by 2k:
(kx)² = 2k(WLsin (Ff)L + ½(W/g)v²)

And from the inequality above,
(kx)² (Fstat_max + Wsin)²

Combine these two to get:
2k(WLsin (Ff)L + ½(W/g)v²) (Fstat_max + Wsin)²

or:
k ½(Fstat_max + Wsin)² / (WLsin (Ff)L + ½(W/g)v²)

Plug in the values and get answer.

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