When a25 kg crate is pushed across a frictionless horizontal floor with a force of 200n directed 20°below the horizontal, magnitude of the normal force of the floor on the crate is

310N

Explanation:

These are the options for the question

A)68N

B. 68N

C. 180N

D. 250N

the normal force of the floor on the crate can be calculated using below formula

N=W+ Fsin(x)

= mg+ Fsin(x)

m= mass of the crate= 25kg

g= acceleration due to gravity= 9.8m/s^2

We were told a force of 200n directed 20°below the horizontal

Which means angle(x)= 20°

We were given horizontal floor with a force of 200N I.e F= 200N

If we substitute these values into the formula above we have

N=mg+ Fsin(x)

= (25kg×9.8) + 200sin(20)

=313.5N

Hence, the the magnitude of the normal force of the floor on the crate is approximately 310N

CHECK THE ATTACHMENT FOR THE DIAGRAM

carla slowed down in the beginning of her trip, stopped, and then sped up