Question

A 22.4 kg block is placed on a ramp that makes an angle of 22 ° to the horizontal. The coefficient of friction between the block and the ramp is 0.11. The block is released from rest and allowed to slide down the ramp. What is the acceleration of the block?

Answer 1

Answer: 2.67 m/s2

Step-by-step explanation:

W = mg = (22.4 kg)(9.8 m/s2) = 219.52 N

Wx = (219.52 N)(sin 22) = 82.2 N

Wy = (219.52 N) (cos 22) = 203.5 N

Wy = N (normal force on the block) = 203.5 N

Ff (force of friction) = (coeff. friction)(N) = (0.11)(203.5 N) = 22.4 N

Sum the forces in the x-direction (direction of motion down the ramp):

Fnet = Wx - Ff = 82.2 N - 22.4 N = 59.8 N

a = F/m = 59.8 N / 22.4 kg = 2.7 m/s2