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