Question

A 20 kg wagon is rolling to the right across a floor. A person attempts to catch and stop the crate and applies a force of 70 N, 180.0 on it. If the coefficient of friction is 0.18, calculate the deceleration rate of the wagon as it is caught.

Answer 1

Answer:

1.736m/s²

Explanation:

According to Newton's second law;

$\sum F_x = ma_x$

$Fm - Ff = ma_x$ where;

Fm is the moving force = 70.0N

Ff is the frictional force acting on the body

$Ff = \mu R\\Ff = \mu mg$

$\mu$ is the coefficient of friction

m is the mass of the object

g is the acceleration due to gravity

a is the acceleration/deceleration

The equation becomes;

$Fm - Ff = ma_x\\Fm - \mu mg = ma$

Substitute the given parameters

$Fm - \mu mg = ma\\70 - 0.18(20)(9.8) = 20a\\70-35.28 = 20a\\34.72 = 20a\\a = \frac{34.72}{20}\\a = 1.736m/s^2$

Hence the deceleration rate of the wagon as it is caught is 1.736m/s²