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A truck of mass 2.1 × 10³ kg tows a car of mass 1.3 × 10³ kg along a horizontal road - AQA - A-Level Physics - Question 24 - 2022 - Paper 1

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A truck of mass 2.1 × 10³ kg tows a car of mass 1.3 × 10³ kg along a horizontal road. The total resistive force on the car is 1100 N. The acceleration of the car and... show full transcript

Worked Solution & Example Answer:A truck of mass 2.1 × 10³ kg tows a car of mass 1.3 × 10³ kg along a horizontal road - AQA - A-Level Physics - Question 24 - 2022 - Paper 1

Step 1

Calculate the Net Force on the System

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Answer

To find the tension in the tow rope, we first need to determine the net force acting on the combined system of the truck and the car. The net force ( F_{net} ) is given by:

Fnet=(mtruck+mcar)×aF_{net} = (m_{truck} + m_{car}) \times a

Substituting the values:

  • Mass of the truck, mtruck=2.1×103kgm_{truck} = 2.1 \times 10^3 \, kg
  • Mass of the car, mcar=1.3×103kgm_{car} = 1.3 \times 10^3 \, kg
  • Acceleration, a=2.3m/s2a = 2.3 \, m/s^2

Thus,

Fnet=(2.1×103+1.3×103)×2.3F_{net} = (2.1 \times 10^3 + 1.3 \times 10^3) \times 2.3 Fnet=3.4×103×2.3=7820NF_{net} = 3.4 \times 10^3 \times 2.3 = 7820 \, N

Step 2

Determine the Total Resistance Force

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Next, we must account for the total resistive forces acting on the car. Given that the total resistive force is 1100 N, the tension in the tow rope also needs to overcome this force for the car to accelerate. Thus, we summarize the forces acting on the car:

TFresistive=mcar×aT - F_{resistive} = m_{car} \times a

Where:

TT = tension in the rope

Fresistive=1100NF_{resistive} = 1100 \, N

mcar=1.3×103kgm_{car} = 1.3 \times 10^3 \, kg

a=2.3m/s2a = 2.3 \, m/s^2

Rearranging gives: T=mcar×a+FresistiveT = m_{car} \times a + F_{resistive}

Step 3

Final Calculation for Tension

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Answer

Substituting the known values:

T=(1.3×103×2.3)+1100T = (1.3 \times 10^3 \times 2.3) + 1100 T=2990+1100=4090NT = 2990 + 1100 = 4090 \, N

Rounding to the nearest hundred, the tension in the tow rope is approximately 4100 N.

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