Question

unit exam - forces
a 52 kg mass is pushed across a carpeted floor with a force of -365 n. there is a +53.0 n force due to friction. what is the acceleration of the mass?
a = ? m/s²
hint: f_net = m·a
remember to indicate positive (+) or negative (-) acceleration.
acceleration (m/s²)

Explanation:

Step1: Calculate the net force

The applied force is $F_{applied} = -365\ N$ and the frictional force is $F_{friction} = +53.0\ N$. The net force $F_{net}$ is the sum of these forces:
$$F_{net} = F_{applied} + F_{friction}$$
$$F_{net} = -365\ N + 53.0\ N = -312\ N$$

Step2: Use Newton's second law to find acceleration

Newton's second law is $F_{net} = m \cdot a$, where $m = 52\ kg$. We can solve for $a$:
$$a = \frac{F_{net}}{m}$$
Substitute $F_{net} = -312\ N$ and $m = 52\ kg$:
$$a = \frac{-312\ N}{52\ kg} = -6.0\ m/s^2$$

Answer:

$-6.0$