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a tow truck tows a 1550 kg car along a level road. the cable attactching the car to the tow truck makes an angle of 37 above the horizontal. the coefficient of friction between the car and the road is u=0.198 and the tension in the cable is 6510 N. Determine the magnitude of the acceleration of the car.

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asuwafohjames

The acceleration of the car is 1.4 m/s².

What is acceleration?

This can be defined as the rate of change of velocity.

To calculate the acceleration of the car, we use the formula below.

Formula:

  • Tcos∅ = ma+mgμ............. Equation 1

Where:

  • T = Tension in the cable
  • m = mass of the car
  • a = acceleration of the car
  • g = acceleration due to gravity
  • ∅ = angle of the cable to the horizontal
  • μ = coefficient of friction.

Make a the subject of the equation.

  • a = (Tcos∅-mgμ)/m............ Equation 2

From the question,

Given:

  • T = 6510 N
  • m = 1550 kg
  • ∅ = 37°
  • g = 9.8 m/s²
  • μ = 0.198

Substitute these values into equation 2

  • a = [(6510cos37)-(1550×9.8×0.198)]/1550
  • a = (5199.12-3007.62)/1550
  • a = 2191.5/1550
  • a = 1.4 m/s².

Hence, the acceleration of the car is 1.4 m/s².

Learn more about acceleration here: https://brainly.com/question/605631

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