Answer:
the truck and the car move 586.8 m
Step-by-step 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