Question

A 800kg car is being towed by a 1200kg car using a tow rope. The heavier car accelerates at 2
m/s2.
A) What force is the heavier car producing to pull the lighter car? (assume there are no other
forces except the tension and the force produced by the heavier car’s engine)
B) What is the tension in the rope?
C) If the rope is 2m long made of steel (Young’s modulus = 210GPa) 1cm diameter, how much
will the rope be stretched during the acceleration?

Answer 1

PART A:

Force = mass x acceleration

F = (1200 kg) x (2 m/s²)

F = 2400 N ---------- (ANSWER)

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PART B:

Tension = force of heavier car

T = F = 2400 N ---------- (ANSWER)

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PART C:

Area of cross section of rope, A = pi R² = (3.14) x (0.5 x 10^-2 m)² = 7.85 x 10^-5 m²

Therefore, Length stretched is given by:

ΔL = F.L/AY

ΔL = (2400 N x 2m) / (7.85 x 10^-5 m² x 210 x 10⁹ Pa)

ΔL = 2.91 x 10^-4 m---------- (ANSWER)

or ΔL = 0.291 mm---------- (ANSWER)