Question

2 mark for review a 1.5 kg block is held at rest near the top of a rough incline that makes an angle of 37 from the horizontal the block is then released and starts moving with a constant acceleration. when the block travels 2.0 m, its velocity is 4.0 m/s. the coefficient of kinetic friction between the surface of the incline and the block is most nearly a 0.20 b 0.25 c 0.40 d 0.75

Explanation:

Step1: Find the acceleration

Use the kinematic equation \(v^{2}=v_{0}^{2}+2ax\).
Given \(v_{0} = 0\ m/s\), \(v = 4.0\ m/s\), \(x=2.0\ m\).
Substitute into the equation: \(4^{2}=0 + 2a\times2\).
Simplify: \(16 = 4a\), so \(a=\frac{16}{4}=4\ m/s^{2}\).

Step2: Analyze the forces

The forces along the incline: \(mg\sin\theta - f=ma\), where \(f=\mu_{k}N\).
The normal force \(N = mg\cos\theta\).
So \(mg\sin\theta-\mu_{k}mg\cos\theta=ma\).
Divide both sides by \(m\): \(g\sin\theta-\mu_{k}g\cos\theta=a\).
Given \(g = 10\ m/s^{2}\), \(\theta = 37^{\circ}\), \(a = 4\ m/s^{2}\).
\(\mu_{k}=\frac{g\sin\theta - a}{g\cos\theta}\).
\(\sin37^{\circ}\approx0.6\), \(\cos37^{\circ}\approx0.8\).
Substitute: \(\mu_{k}=\frac{10\times0.6 - 4}{10\times0.8}=\frac{6 - 4}{8}=\frac{2}{8}=0.25\).

Answer:

B. \(0.25\)