A 6 kg block is released from rest at the top of an incline, as shown above, and slides to the bottom. The incline is 1.97 m long, is inclined at an angle of 29.4°, and the coefficient of kinetic friction between the block and the incline is 0.21. a. Calculate the normal force the incline puts on the block.

The net force on the block acting perpendicular to the incline is

∑ F = n - w cos(29.4°) = 0

where n is the magnitude of the normal force and w = m g is the weight of the block.

The equation itself comes from splitting up the forces acting on the block into components pointing parallel or perpendicular to the incline. The only forces acting on the block in the perpendicular direction are the normal force and the perpendicular component of the block's weight.

Solve for n :

n = m g cos(29.4°)

n = (6 kg) (9.80 m/s²) cos(29.4°)

n ≈ 51.2 N

answer: the answer is a

the answer is b

explanation: