A car, towing a trailer, starts to move and after 12s reaches a speed of 15m / s.

| January 9, 2021 | education

A car, towing a trailer, starts to move and after 12s reaches a speed of 15m / s. The drag coefficient is 0.3. Determine the elastic strength of the adhesion.

V0 = 0 m / s.
V = 15 m / s.
g = 9.8 m / s2.
m = 5 t = 5000 kg.
t = 12 s.
μ = 0.3.
Fupr -?
Let us write Newton’s 2 law: m * a = Fcont – Fcopr, where m is the body mass, a is the body’s acceleration, Fcr is the elastic force of adhesion, Fcopr is the resistance force.
Fcont = m * a + Fc.
The resistance force Fcopr is expressed by the formula: Fcopr = μ * m * g.
The acceleration of the body a is expressed by the formula: a = (V – V0) / t. Since the car begins to move from a state of rest, the formula will take the form: a = V / t.
Fcont = m * V / t + μ * m * g = m * (V / t + μ * g).
Fcont = 5000 kg * (15 m / s / 12 s + 0.3 * 9.8 m / s2) = 20950 N.
Answer: the elastic force of adhesion is Fel = 20950 N.



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