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
