Question

1. alistair places a 6.40 kg block on an inclined plane that is 23° above the horizontal. the block starts from rest, and in 2.20 seconds, travels 1.40 m. what is the force of friction on the block? 19.5 n 20.8 n 3.70 n 24.5 n

Explanation:

Step1: Calculate the acceleration

Use the kinematic equation \(s = v_0t+\frac{1}{2}at^{2}\). Since \(v_0 = 0\) (starts from rest), we have \(s=\frac{1}{2}at^{2}\). Then \(a=\frac{2s}{t^{2}}\).
Substitute \(s = 1.40\space m\) and \(t = 2.20\space s\) into the formula:
\(a=\frac{2\times1.40}{(2.20)^{2}}=\frac{2.80}{4.84}\approx0.58\space m/s^{2}\)

Step2: Analyze the forces along the incline

The force along the incline due to gravity is \(F_{g\parallel}=mg\sin\theta\), where \(m = 6.40\space kg\), \(g = 9.8\space m/s^{2}\), and \(\theta=23^{\circ}\).
\(F_{g\parallel}=6.40\times9.8\times\sin(23^{\circ})\)
\(\sin(23^{\circ})\approx0.391\), so \(F_{g\parallel}=6.40\times9.8\times0.391 = 6.40\times3.832\approx24.5\space N\)

According to Newton's second law \(F_{net}=ma\), and \(F_{net}=F_{g\parallel}-F_f\) (where \(F_f\) is the frictional force)
\(F_f=F_{g\parallel}-ma\)
Substitute \(m = 6.40\space kg\), \(a\approx0.58\space m/s^{2}\), and \(F_{g\parallel}\approx24.5\space N\)
\(F_f = 24.5-6.40\times0.58=24.5 - 3.712\approx20.8\space N\)

Answer:

20.8 N