Question

A 57 kg boy and a 41 kg girl use an elastic rope while engaged in a tug-of-war on a frictionless icy surface. If the acceleration of the girl toward the boy is 2.6 m/s 2 , determine the magnitude of the acceleration of the boy toward the girl. Answer in units of m/s 2 .

Answer 1

Answer:

Acceleration of the boy a₁:

$a_{1} = 1.87 \frac{m}{s^{2} }$

Explanation:

Conceptual analysis

We apply Newton's second law to the boy and the girl:

F = m*a (Formula 1)

F : Force in Newtons (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Nomenclature

m₁ : boy mass

m₂ :  girl mass

a₁ : boy acceleration

a₂ :  girl acceleration

F₁ : boy acceleration

F₂ :  girl acceleration

Known data

m₁ =  57 kg

m₂ =  41  kg

a₂ = 2.6 m/s²

Problem development

We apply to Newton's third law of action and reaction, then:

F₁ = F₂ , We apply the formula (1):

m₁*a₁ = m₂*a₂

$a_{1} = \frac{m_{2}* a_{2} }{m_{1} }$

$a_{1} = \frac{41* 2.6 }{ 57 }$

$a_{1} = 1.87 \frac{m}{s^{2} }$