Question

problem 4: a frustrated student pushes a calculator across the table. find the acceleration of the calculator as it slows down before falling off the edge, given a coefficient of kinetic friction of 0.16 and a calculator mass of 25 g.

Explanation:

Step1: Identify forces acting

The frictional force $F_f=\mu_k F_N$. On a flat - surface, $F_N = mg$.

Step2: Calculate normal force

First, convert mass to SI unit: $m = 25\ g=0.025\ kg$. The normal force $F_N=mg$, where $g = 9.8\ m/s^2$. So $F_N=(0.025\ kg)\times(9.8\ m/s^2)=0.245\ N$.

Step3: Calculate frictional force

The frictional force $F_f=\mu_k F_N$. Given $\mu_k = 0.16$, then $F_f=(0.16)\times(0.245\ N)=0.0392\ N$.

Step4: Use Newton's second law

According to Newton's second law $F = ma$, and the only horizontal force acting on the calculator as it slows down is the frictional force, so $F_f=ma$. Then $a=\frac{F_f}{m}$.

Step5: Calculate acceleration

Substitute $F_f = 0.0392\ N$ and $m = 0.025\ kg$ into the acceleration formula: $a=\frac{0.0392\ N}{0.025\ kg}=1.568\ m/s^2$. The acceleration is negative because it is a deceleration, so $a=- 1.568\ m/s^2$.

Answer:

$-1.57\ m/s^2$ (rounded to two decimal places)