Problem — Block pushed with friction

Problem

A block of mass $m = 3{,}0\;\mathrm{kg}$ is pushed across a horizontal surface by a constant force $F = 20\;\mathrm{N}$ parallel to the surface, over a distance $d = 2{,}0\;\mathrm{m}$. The coefficient of kinetic friction is $\mu_d = 0{,}30$. Using the energy balance, determine the block’s final speed, starting from rest.

Solution

Energy balance. The work of the applied force feeds both the block’s kinetic energy and the energy dissipated by friction: $L_F = \Delta E_k + L_\text{attr}$ Friction force. On the horizontal surface the normal force balances the weight, $N = m\,g$, so $F_a = \mu_d\,N = \mu_d\,m\,g = 0{,}30\cdot 3{,}0\;\mathrm{kg}\cdot 9{,}8\;\mathrm{m/s^2} = 8{,}82\;\mathrm{N}$ Final kinetic energy. Starting from rest, $\Delta E_k = \tfrac{1}{2}mv^2$, and the net work is $\frac{1}{2}\,m\,v^2 = (F - F_a)\,d = (20 - 8{,}82)\;\mathrm{N}\cdot 2{,}0\;\mathrm{m} = 22{,}4\;\mathrm{J}$ Solving for $v$: $v = \sqrt{\frac{2\cdot 22{,}4\;\mathrm{J}}{3{,}0\;\mathrm{kg}}} = \sqrt{14{,}9}\;\mathrm{m/s}$ $\ev{v \approx 3{,}9\;\mathrm{m/s}}$

Links

Topics: Work and energy Concepts: Thermal energy from friction · Kinetic friction Skills: Bubble-diagram energy balance Objects: Block