Problem — Pushing a moving crate

Problem

Assuming the crate is already moving, push it — mass $30\;\mathrm{kg}$ — for $5{,}0\;\mathrm{m}$ along the floor, applying a horizontal force of $18\;\mathrm{N}$. The coefficient of kinetic friction is $\mu_d = 0{,}10$. Find: (a) the work done by you; (b) the work done by friction; (c) the change in kinetic energy.

Solution

(a) Work of the push. Force parallel to the displacement ($\theta = 0$): $L_\text{spinta} = F\,d = 18\;\mathrm{N}\cdot 5{,}0\;\mathrm{m} = 90\;\mathrm{J}$ $\ev{L_\text{spinta} = 90\;\mathrm{J}}$

(b) Work of friction. The friction force is $F_a = \mu_d\,m\,g = 0{,}10\cdot 30\;\mathrm{kg}\cdot 9{,}8\;\mathrm{m/s^2} = 29{,}4\;\mathrm{N}$, opposite to the displacement ($\theta = 180^\circ$), so its work is negative: $L_\text{attr} = -F_a\,d = -29{,}4\;\mathrm{N}\cdot 5{,}0\;\mathrm{m} = -147\;\mathrm{J}$ $\ev{L_\text{attr} \approx -1{,}5\cdot 10^2\;\mathrm{J}}$

(c) Change in kinetic energy. By the work–energy theorem, $\Delta E_\text{cin}$ is the total work: $\Delta E_\text{cin} = L_\text{spinta} + L_\text{attr} = 90\;\mathrm{J} - 147\;\mathrm{J} = -57\;\mathrm{J}$ $\ev{\Delta E_\text{cin} \approx -57\;\mathrm{J}}$ The kinetic energy decreases: the $18\;\mathrm{N}$ push is not enough to overcome the $29{,}4\;\mathrm{N}$ friction, so the crate slows down.

Collegamenti

Argomenti: Lavoro ed energia Concetti: Lavoro di una forza · Energia termica da attrito · Attrito dinamico Competenze: Bilancio energetico a bolle Oggetti: Blocco