Question

1. the force of gravity on a wooden crate is 500.0 n. if it requires 1.5 x 10³ j of work to push the crate at a steady speed along the floor for a distance of 5.0 m, what is the coefficient of friction between the floor and the crate?

Explanation:

Step1: Recall work - force - distance formula

Work \(W = Fd\), where \(W\) is work, \(F\) is force and \(d\) is distance. We know \(W=1.5\times 10^{3}\text{ J}\) and \(d = 5.0\text{ m}\), so the applied force \(F_{applied}=\frac{W}{d}\).

$$F_{applied}=\frac{1.5\times 10^{3}\text{ J}}{5.0\text{ m}}=300\text{ N}$$

Step2: Since the crate moves at a steady speed

The net force on the crate is zero (\(\sum F = 0\)). The frictional force \(F_f\) is equal to the applied force \(F_{applied}\) because \(F_{net}=F_{applied}-F_f = 0\) (Newton's first - law for constant velocity motion), so \(F_f=300\text{ N}\).

Step3: Recall the formula for frictional force

The normal force \(N\) on the crate is equal to the gravitational force \(F_g\) acting on it since the crate is on a horizontal surface (\(N = F_g\)). Given \(F_g=500\text{ N}\), and the formula for frictional force is \(F_f=\mu N\), where \(\mu\) is the coefficient of friction. We can solve for \(\mu\): \(\mu=\frac{F_f}{N}\).

$$ \mu=\frac{300\text{ N}}{500\text{ N}} = 0.6$$

Answer:

0.6