Accelerations at fixed force

Problem

Accelerations at fixed force. Four crates of different mass receive the same horizontal push $F = 30\;\text{N}$ on a smooth floor: (A) $m = 2{,}0\;\text{kg}$; (B) $m = 5{,}0\;\text{kg}$; (C) $m = 10\;\text{kg}$; (D) $m = 30\;\text{kg}$. Rank the accelerations from largest to smallest.

Solution

On the smooth floor the only horizontal force is the push $F$, the same for all. From the second law $a = F/m$: for equal force, the acceleration is inversely proportional to the mass. The lightest crate accelerates the most. Let us calculate:

$a_A = \frac{30}{2{,}0} = 15\;\text{m/s}^2 \qquad a_B = \frac{30}{5{,}0} = 6{,}0\;\text{m/s}^2$

$a_C = \frac{30}{10} = 3{,}0\;\text{m/s}^2 \qquad a_D = \frac{30}{30} = 1{,}0\;\text{m/s}^2$

Ranking from largest to smallest (increasing masses → decreasing accelerations):

$\ev{a_A > a_B > a_C > a_D \quad (15,\; 6,\; 3,\; 1\;\text{m/s}^2)}$

Links

Topics: Dynamics Concepts: Newton’s second law Skills: Ranking reasoning