A car of mass 1357.7 kg rolls down a sloping road when the handbrake fails. The road slopes at an angle of 5.3 degrees to the horizontal, and the frictional forces on the car add up to 168.1 newton.What is the net force acting on the car?

Question

BlackTom AI · Accepted Answer

The component of the car's weight down the slope is mg sinθ, so net force along the slope = (weight component down-slope) − (frictional forces).

1) Relevant concepts:
- Weight = mg acts vertically downward.
- The component of weight parallel to an incline of angle θ is mg sinθ (driving the car down).
- Friction opposes motion along the slope; net force = mg sinθ − friction.

2) Step-by-step reasoning and calculations:
- Mass m = 1357.7 kg, g = 9.81 m/s², θ = 5.3°, friction = 168.1 N.
- Compute the weight: mg = 1357.7 × 9.81 = 13,320.037 N (intermediate value).
- Compute sin(5.3°). Using a calculator gives sin(5.3°) ≈ 0.0923703.
- Weight component down the slope: mg sinθ = 13,320.037 × 0.0923703 ≈ 1,230.29 N (intermediate multiplication shown).
- Subtract friction to get the net force: Fnet = 1,230.29 − 168.1 = 1,062.19 N.

3) Intermediate steps shown:
- mg = 1357.7 × 9.81 = 13,320.037 N
- sin(5.3°) ≈ 0.0923703
- mg sinθ ≈ 13,320.037 × 0.0923703 = 1,230.29 N
- Fnet = 1,230.29 − 168.10 = 1,062.19 N

4) Why this is correct:
- The gravity component along the slope is the only driving force; friction opposes it. Subtracting the known friction from mg sinθ gives the unbalanced (net) force causing acceleration down the slope.

5) Why the other options are incorrect:
- a. 13430.19 N is far larger (close to mg with wrong rounding or using g=9.9) and ignores trig decomposition and friction.
- c. 13093.99 N and d. 1398.39 N arise from using incorrect components (for example using mg cosθ or wrong subtraction/addition) or arithmetic mistakes; they do not correctly compute mg sinθ minus friction.

6) Conclusion:
The correct net force down the slope is 1,062.19 N, which matches the provided answer (1062.19 N).

A car of mass 1357.7 kg rolls down a sloping road when the handbrake fails. The road slopes at an angle of 5.3 degrees to the horizontal, and the frictional forces on the car add up to 168.1 newton.What is the net force acting on the car?数值题

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