A car of mass 1344 kg rolls down a sloping road when the handbrake fails. The road slopes at an angle of 6 degrees to the horizontal, and the frictional forces on the car add up to 253 newton.What is the car's acceleration down the slope? Give your answer to one decimal place.

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The scenario involves a car sliding down a slope with frictional forces opposing motion. We start by identifying the given quantities: mass m = 1344 kg, slope angle θ = 6°, gravitational acceleration g ≈ 9.8 m/s^2, and total frictional force F_f = 253 N along the plane.

First, calculate the component of gravity driving the car down the slope. This is given by m g sin(θ). Using sin(6°) ≈ 0.104528, we have:
- g sin(θ) ≈ 9.8 × 0.104528 ≈ 1.0244 m/s^2
- m g sin(θ) ≈ 1344 × 1.0244 ≈ 1376.8 N
This is the downhill force acting along the slope.

Next, subtract the frictional forces that oppose the motion to obtain the net force along the slope:
- F_net = m g sin(θ) − F_f ≈ 1376.8 N − 253 N ≈ 1123.8 N

Now apply Newton’s second law along the slope to find the acceleration: a = F_net / m.
- a ≈ 1123.8 N / 1344 kg ≈ 0.835 m/s^2

Rounding to one decimal place (as requested) gives a ≈ 0.8 m/s^2.

If one were to check the logic, the key steps are (1) resolve gravity into a downslope component, (2) subtract opposing friction, (3) divide by the mass to get acceleration. Any misstep would typically involve miscomputing sin(θ) or mixing up the direction of friction, which would yield an incorrect sign or magnitude. Ensure units are consistent throughout: forces in newtons and mass in kilograms lead to acceleration in m/s^2.

MUF0121 Physics Unit 1 - Semester 2, 2025 5.8 Application of forces

A car of mass 1344 kg rolls down a sloping road when the handbrake fails. The road slopes at an angle of 6 degrees to the horizontal, and the frictional forces on the car add up to 253 newton.What is the car's acceleration down the slope? Give your answer to one decimal place.

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