Answer:
1.736m/s²
Explanation:
According to Newton's second law;
[tex]\sum F_x = ma_x\\[/tex]
[tex]Fm - Ff = ma_x\\[/tex] where;
Fm is the moving force = 70.0N
Ff is the frictional force acting on the body
[tex]Ff = \mu R\\Ff = \mu mg\\[/tex]
[tex]\mu[/tex] 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;
[tex]Fm - Ff = ma_x\\Fm - \mu mg = ma\\[/tex]
Substitute the given parameters
[tex]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\\[/tex]
Hence the deceleration rate of the wagon as it is caught is 1.736m/s²
