a block sliding on ground where 0.243 experiences a 24.7 friction force. what is the normal force acting on the block?

The normal force acting on the block is 101.64N.

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west