Question

A 9.00-kg block slides down a frictionless plane having an incline of 27.0 The block starts from rest from the top of the incline, and the length of the incline is 2.00 m. Draw a free-body diagram for the block. Determine the normal force acting on the block Answer: Check If the block starts from rest, determine the speed of the block at the bottom of the incline. (Hint: determine the acceleration of the block down the incline, then use the kinematic equations.) Answer

Answer 1

Given, m = 9 kg theta = 27 degree length of incline (L) = 2 m Normal force acting on the block: therefore N = mg cos theta = (9)(9.8) cos (27) therefore N = 78.59 N Here, acceleration of block: a = F_net/m = mg sin theta/m = g sin theta therefore a = (9.8) sin(27) = 4.45 m/s^2 and, s = 2 m Using formula: V^2 - U^2 = 2 as therefore V = squareroot 2 as = squareroot 2(4.45)(2) = 4.22 m/s therefore V = 4.22 m/s speed of the block at the bottom of the incline.