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fiasKO [112]
3 years ago
14

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 acceleratio

n 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 .
Physics
1 answer:
Sergeeva-Olga [200]3 years ago
5 0

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} }

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8)Acceleration of the car is 11.17 m/s².

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Explanation:

8) Acceleration of the car is determined by finding the ratio of change in velocity to the time taken for the change in velocity. As here the initial velocity is 18.5 m/s and the final velocity is 46.1 m/s, then in 2.47 seconds, the acceleration is

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Average acceleration = (7.5-5)/1.25=2 m/s².

So the average acceleration of the sprinter is 2 m/s².

12)Since the acceleration of the stroller is given as 0.6 m/s². And the initial velocity is given as zero. So for the time of 4.75 minutes, the velocity will be equal to the final velocity.

As per equations of motion,

v = u +at

As u =0, a = 0.6 m/s² and t = 4.75 ×60 s = 285 s

So v = 0.6×285 = 171 m/s

Thus, the velocity of the stroller after 4.75 minutes is 171 m/s.

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Then v = u+at=0+(2.2×5)=11 m/s

So the skier will be moving as 11 m/s speed.

14) Here a = 50 m/s² and v = 28 m/s with u = 0 so time taken to reach the speed of 28 m/s is

v = u +at = 0+ (50 t)

28 = 50 t

t = 28/50 = 0.56 s

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