Ablock with mass m = 6 kg is sitting on a horizontal surface and not moving. the free-fall acceleration is g = 9.81 m/s2. answer the following questions.
a. write an expression for the magnitude of the force of gravity fg on the block.
b. calculate the magnitude of the force of gravity fg on the block in newtons.
c. in what direction is the force of gravity in this problem?
d. what is the magnitude of the normal force fn in newtons?
e. in what direction does the normal force act?

a) Fg = m* g

b) Fg = 58.86 N

c) pointing down (towards the center of the Earth)

d) N = 58.86 N

e) pointing up

Explanation:

a) Force is defined as mass of the object times the acceleration it is subjected to. In our case, the block is subjected to the acceleration of gravity (g), therefore the force due to gravity (Fg) is Fg=m*g

b) perform the indicated product of part a):

Fg = m*g = 6 kg * 9.81 m/s^2 = 58.86 N

c) since the object is not moving, that means that the contact force exerted by the surface were the block is resting, must equal the acting force of gravity, but pointing in opposite direction, so the NET force on the block is zero. Therefore, the magnitude of the normal force (N) must equal the magnitude of the object's weight: N = 58.86 N

d) As explained in part c), the normal force must be acting in opposite direction to the force of gravity, which means pointing up.

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