Wb = M*g = 1.70 * 9.8 = 16.7 N.
Fp = 16.7*sin27.4 = 7.67 N. = Force parallel to the incline.
a. a = Fp/M = 7.67/1.70 = 4.51 m/s^2.
b. PE = Mg*h = 16.7*0.485 = 8.08 J.
KE = 0.5M*V^2 = 8.08.
0.85*V^2 = 8.08.
V^2 = 9.51.
V = 3.08 m/s.
c.
A block of mass m = 1.70kg is released from rest h = 0.485m from the surface of a table, at the top of a θ = 27.4° incline as shown in the figure below.
I got a few of the questions
a)The frictionless incline is fixed on a table of height H = 1.98m. Determine the acceleration of the block as it slides down the incline.
answer:4.51 m/s^2
b) What is the velocity of the block as it leaves the incline?
answer: 3.08 m/s
questions need help with
c) How far from the table will the block hit the floor?
d) How much time has elapsed between when the block is released and when it hits the floor?
1 answer