g=9.8m/s2A block of mass m = 18 kg accelerates across a rough table when a force F = 100 N acts upon it at an angle = 35° with the horizontal. The force of friction on the block is f = 33 N.(a) What is the acceleration of the block? m/s2(b) What is the coefficient of friction of the block?

The free body diagram of the block can be shown as,

Part (a)

According to free body diagram, the net force which acts along x-axis is given as,

According to Newton's law, the net force acting along x-axis is,

Plug in the known value,

Substitute the known values,

Thus, the acceleration of the block is 2.72 m/s2.

Part (b)

According to free body diagram, the net force acting along y-axis is,

Since the block accelerates along x-axis therefore, the net force along y-axis is zero which can be expressed as,

Substitute the known values,

The formula to calculate the frictional force is,

Substitute the known values,

Thus, the coefficient of friction of block is 0.277.

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this