Question

a 410 - n rightward force is applied to a 62 - kg object. there is 193 n of friction. determine the acceleration. f_grav = mg=(62 kg)(9.8 n/kg) f_grav = 607.6 n f_app = 410 n f_norm = 607.6 n the f_net is the vector sum of all the forces. f_net = 410 n - 193 n = 217 n a = f_net/m a = 217 n/62 kg calculate the acceleration.

Explanation:

Step1: Calculate net - force

The net force $F_{net}$ is the vector sum of all forces. Given an applied right - ward force of $410\ N$ and a frictional force of $193\ N$, so $F_{net}=410\ N - 193\ N=217\ N$.

Step2: Use Newton's second law

Newton's second law is $F = ma$, where $F = F_{net}$, $m = 62\ kg$ and we want to find $a$. Rearranging for $a$ gives $a=\frac{F_{net}}{m}$.

Step3: Calculate acceleration

Substitute $F_{net}=217\ N$ and $m = 62\ kg$ into the formula $a=\frac{F_{net}}{m}=\frac{217\ N}{62\ kg}=3.5\ m/s^{2}$.

Answer:

$3.5\ m/s^{2}$