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

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 550-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 8.0 m along the ramp. Once stopped, a crate must not rebound back up the ramp. Calculate the force constant of the spring that will be needed in order to meet the design criteria.

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Tài Đức 3 years 2021-08-14T18:08:35+00:00 1 Answers 233 views 0

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  1. Nho
    0
    2021-08-14T18:09:46+00:00
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    Answer:

    The force constant ,I = 2394N/m

    Explanation:

    Given:

    Weight of crate,Wg = 1470N

    Theta = 22.0°

    Kinetic friction,Fk= Fs(max) = 550N

    Total length of ramp =8.0m

    If y =0 at the bottom of the ramp

    y1 = d Sin theta

    y1 = 8 × Sin 22°

    y1 = 3.0m

    y2 = 0

    V1=1.8m/s

    V2 = 0

    The equation combining the gravitational and elastic potential energy, and the work energy theorem is given by:

    K1 + Ugrav1 + Uel1 + W = K2 + Uel1 ..eq1

    Where KE is given by: 1/2mv^2

    Gravitational potential energy is given by: Ugrav = mgy …eq2

    Elastic potential energy = Uel = 1/2Kx^2 ..eq3

    The restoring force constant of a spring compressed by a distance x is given by:

    Fx = Kx ..eq4

    Workdone by a constant force is given by W = Fscostheta …eq5

    Where s = displacement

    Theta = the angle between the force and the displacement

    Work,W = Wf = 550 ×8 × cos 180°

    W = -4400J

    From the weight of the crate.mass is:

    Wg/g = m

    1470/9.8 = 150kg

    K1 = 1/2 ×150×1.8^2 = 243m/s

    The crate comes to rest at K2=0

    Ugrav1 = 150 × 9.8 × 3 = 4410J

    Ugrav2 = 150 × 9.8 x 0 = 0J

    Uel1= 0 Spring at equilibrium

    Substituting the values of the energies and work

    253 + 4410 + 0 – 4400 = 0 + 0 + Uel1

    Uel1 = 253J

    Substituting into eq3

    253 = 1/2 Kx^2 = 1/2 Kx(x)

    Kx = 506/x

    Since crate remains at rest,we use Newton’s 2nd law

    Fx = fs + Wsin theta

    Fx = 550 + 1470 sin 22

    Fx = 1100.7N

    Substituting into eq4

    Kx = 1100.7

    X = 506/1100.7 = 0.46m

    Kx = 1100.7

    K = 1100.7/0.47

    K = 2394N/m

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