Question

1. a force of 40 n accelerates a 5 kg block a6 6.0 m/s2 along a horizontal surface.

a) what would be the blocks acceleration if the surface was frictionless?
b) how large is the kinetic friction force?
c) what is the coefficient of kinetic friction?

Explanation:

Step1: Apply Newton's second law for frictionless case

Newton's second law is \(F = ma\). Given \(F = 40\ N\) and \(m = 5\ kg\), we solve for \(a\).

$$a=\frac{F}{m}$$

$$a=\frac{40}{5}$$

$$a = 8\ m/s^{2}$$

Step2: Find the net force in the original case

Using \(F_{net}=ma\), with \(m = 5\ kg\) and \(a=6.0\ m/s^{2}\), \(F_{net}=5\times6 = 30\ N\)

Step3: Calculate the kinetic friction force

The applied force \(F = 40\ N\) and \(F_{net}=F - F_{f}\) (where \(F_{f}\) is the friction force). So \(F_{f}=F - F_{net}\)

$$F_{f}=40 - 30$$

$$F_{f}=10\ N$$

Step4: Find the coefficient of kinetic friction

The normal force \(N = mg\) (since on a horizontal surface), \(m = 5\ kg\) and \(g = 10\ m/s^{2}\), so \(N=5\times10 = 50\ N\). The formula for kinetic friction is \(F_{f}=\mu_{k}N\), solving for \(\mu_{k}\) gives \(\mu_{k}=\frac{F_{f}}{N}\)

$$ \mu_{k}=\frac{10}{50}$$

$$ \mu_{k}=0.2$$

Answer:

a) \(8\ m/s^{2}\)
b) \(10\ N\)
c) \(0.2\)