Question

4 a 1.45 kg box is placed on a ramp that is 21° above the horizontal. if the coefficient of static friction is 0.49, and the coefficient of kinetic friction is 0.32, what will happen to the box?
the box will not move.
the box will accelerate at 0.58 m/s²
the box will accelerate at 0.47 m/s²
the box will accelerate at 0.22 m/s²

Explanation:

Step1: Calculate the maximum static - friction force

The normal force \(N = mg\cos\theta\), where \(m = 1.45\space kg\), \(g=9.8\space m/s^{2}\), \(\theta = 21^{\circ}\).
The maximum static - friction force \(f_{s,\text{max}}=\mu_{s}N=\mu_{s}mg\cos\theta\)
Substitute \(\mu_{s}=0.49\), \(m = 1.45\space kg\), \(g = 9.8\space m/s^{2}\), \(\theta=21^{\circ}\)
\(N=1.45\times9.8\times\cos(21^{\circ})\approx1.45\times9.8\times0.934 = 13.39\space N\)
\(f_{s,\text{max}}=0.49\times13.39\approx6.56\space N\)

The gravitational force along the incline \(F_{g\parallel}=mg\sin\theta\)
Substitute \(m = 1.45\space kg\), \(g = 9.8\space m/s^{2}\), \(\theta = 21^{\circ}\)
\(F_{g\parallel}=1.45\times9.8\times\sin(21^{\circ})\approx1.45\times9.8\times0.358=5.11\space N\)
Since \(F_{g\parallel}

Step2: Calculate the kinetic - friction force and acceleration

The kinetic - friction force \(f_{k}=\mu_{k}N\), where \(\mu_{k}=0.32\), \(N = 13.39\space N\) (from Step 1)
\(f_{k}=0.32\times13.39 = 4.29\space N\)
Using Newton's second law \(F_{net}=ma\), and \(F_{net}=mg\sin\theta - f_{k}\)
\(ma=mg\sin\theta - f_{k}\)
\(a=\frac{mg\sin\theta - f_{k}}{m}\)
Substitute \(m = 1.45\space kg\), \(mg\sin\theta = 5.11\space N\), \(f_{k}=4.29\space N\)
\(a=\frac{5.11 - 4.29}{1.45}=\frac{0.82}{1.45}\approx0.565\approx0.58\space m/s^{2}\)

Answer:

The box will accelerate at \(0.58\space m/s^{2}\)