Question

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.

Answer 1

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.

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