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A 4.2 kg object hangs at one end of a rope that is attached to a support on a railroad boxcar. When the car accelerates to the right, the rope makes an angle of 28◦ with the vertical The acceleration of gravity is 9.8 m/s 2 . a 4.2 kg 28 ◦ Find the acceleration of the car.

Respuesta :

Answer:

5.21 m/s²

Step-by-step explanation:

Data provided in the question:

Mass of the hanging object = 4.2 kg

Angle made by the rope with vertical = 28°

Acceleration of gravity = 9.8 m/s²

Now,

let the tension in the rope be 'T'

Thus,

The vertical component of the tension = Tcos(28°)

The horizontal component of the tension = Tsin(28°)

Thus,

For the equilibrium

Net force in vertical direction

Tcos (28) = mg  [as Fynet = 0] ..............(1)

and,

Net Horizontal force

Tsin(28°) = ma [as the acceleration of the boxcar is what is causing the object to make the angle]    ....................(2)

Now, 2 divided by 1

tan(28°) = [tex]\frac{mg}{ma}[/tex]

a = g[tan(28°)]

or

a = 9.8 × tan(28°)

or

a = 5.21 m/s²

fichoh

Using the required equation where, [tex] a = g \times tan θ[/tex], the acceleration of the car is [tex]5.21 \: ms^{-2}[/tex]

Using the relation :

  • [tex] tan θ = \frac{ma}{mg}[/tex]

  • θ = angle of inclination

  • m = mass of the body ; g = acceleration due to gravity = 9.8 m/s²

Making 'a' the subject :

[tex] a = g \times tan θ[/tex]

Plugging in the values :

[tex] a = 9.8 \times tan(28)[/tex]

[tex] a = 9.8 \times 0.5317094 [/tex]

[tex] a = 5.21 \: ms^{-2}[/tex]

Therefore, the acceleration of the car is [tex] 5.21 \: ms^{-2}[/tex]

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