An elastic towrope has spring constant 1300 n/m. it’s connected between a truck and a 1900-kg car. as the truck tows the car, the rope stretches 55 cm. starting from rest, how far do the truck and the car move in 1 min? assume the car experiences negligible friction.

the truck and the car move 586.8 m

Explanation:

Hook's Law: Hook's law state that the force applied to an elastic material is directly proportional to its extension, provided the elastic limit is not exceeded.

From hook's law,

F= ke Equation 1

Where F = applied force, k = force constant of the spring or elastic material, e = extension

Given: k = 1300 N/m, e =55 cm = (55/100) m = 0.55 m.

Substituting these values into Equation 1,

F = 1300 × 0.55

F = 715 N.

But Force (F) = ma Equation 2

making a the subject of the equation,

a = F/m Equation 3

Where m = mass of the car, a = acceleration of the car.

Given: m = 1900 kg, and F = 715 N

Substituting these values into equation 3

a = 715/1900

a = 0.376 m/s²

Using one of the equation of motion,

S = ut + 1/2at² Equation 4

Where S = distance covered, u = initial velocity of the car, t = time, a = acceleration

Given: t= 1 min (1×60) = 60 s, u = 0 m/s ( at rest), a = 0.376 m/s²

Substituting these values into equation 4

S = (0 × 60) + 1/2(0.376×60²)

S = (3600 × 0.376)/2

S = 586.8 m

Therefore the truck and the car move 586.8 m

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distance = 150m

displ = root (50 ^2 + 100^2)

root(2500+10000)

root 125000