Question

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homework 3 begin date: 8/17/2025 11:59:00 pm due date: 9/9/2025 11:59:00 pm end date: 9/9/2025 11:59:00 pm
problem 8: (6% of assignment value)
a toy car rolls down a ramp at a constant velocity. the car’s mass is m = 1.2 kg, and the ramp makes an angle of θ = 11° with respect to the horizontal. assume that the rolling resistance is negligible. note the coordinate system provided on the image where the x axis is parallel to and directed down the incline.

• part (a) ✔️

what is the magnitude, in meters per squared second, of the cars acceleration?
a = 0.000 m/s² ✔️ correct!

• part (b) ✔️

what is the numeric value, in newtons, for the sum of the forces in the x direction?
σ fₓ = 0.000 n ✔️ correct!

• part (c) ✔️

assuming the car experiences only air resistance, with magnitude fᵣ, in opposition to its motion. write an expression for the sum of the forces in the x direction in terms of m, g, θ, and fᵣ. (use the acceleration due to gravity, g, and the variables provided.)
σ fₓ = m g sin(θ) - fᵣ ✔️ correct!

• part (d)

what is the magnitude, in newtons, of the force caused by air resistance, fᵣ? (maintain the assumption that the car’s velocity is constant.)
fᵣ = n

Explanation:

Step1: Recall Newton's First Law (Constant Velocity)

Since the car's velocity is constant, the net force in the x - direction is zero. From part (c), the net force in the x - direction is given by $\sum F_{x}=mg\sin(\theta)-F_{r}$. And since $\sum F_{x} = 0$ (because $a = 0$ from part (a), and $F=ma$), we have $mg\sin(\theta)-F_{r}=0$.

Step2: Solve for $F_{r}$

We can re - arrange the equation $mg\sin(\theta)-F_{r}=0$ to solve for $F_{r}$. Adding $F_{r}$ to both sides and subtracting 0 from both sides (or directly re - arranging), we get $F_{r}=mg\sin(\theta)$.
We know that $m = 1.2\space kg$, $g=9.8\space m/s^{2}$ and $\theta = 11^{\circ}$. First, calculate $\sin(11^{\circ})$. $\sin(11^{\circ})\approx0.1908$.
Then, substitute the values into the formula: $F_{r}=1.2\times9.8\times0.1908$.
First, calculate $1.2\times9.8 = 11.76$. Then, calculate $11.76\times0.1908\approx2.244$.

Answer:

$F_{r}\approx2.24\space N$ (or more precisely, using a calculator for $\sin(11^{\circ})$: $\sin(11^{\circ})\approx0.19080899537$, so $F_{r}=1.2\times9.8\times0.19080899537 = 1.2\times1.87\approx2.244\space N$)