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.

Accepted Solution

Answer:5.21 m/s²Step-by-step explanation:Data provided in the question:Mass of the hanging object = 4.2 kgAngle made by the rope with vertical = 28°Acceleration of gravity = 9.8 m/s²Now,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 equilibriumNet force in vertical directionTcos (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 1tan(28°) = [tex]\frac{mg}{ma}[/tex]a = g[tan(28°)]ora = 9.8 × tan(28°)ora = 5.21 m/s²