Question

A 3.0-kg block sits on top of a 5.0-kg block which is on a horizontal surface. The 5.0-kg block is pulled to the right with a force F⃗ . The coefficient of static friction between all surfaces is 0.60 and the kinetic coefficient is 0.37.

a) What is the minimum value of F

needed to move the two blocks?

b) If the force is 10% greater than your answer for (a), what is the acceleration of each block?

Answer 1

(3 + 5) = 8kg., x 9.8 = normal force of 78.4N.
a) Static friction between 5kg. and surface = (78.4 x 0.60) = 47.04N. force required to start movement.
(43.1 x 1.1) = 51.744 N.
The maximum force that the 3kg. block can stand on it before it slides on the 5kg. block = (3 x 9.8) x 0.60, = 117.64N. So it will slide.
Kinetic friction between 3kg. and 5kg. = (3 x 9.8) x 0.37, = 10.878N.
b) 3kg. acceleration = (10.878/3) = 3.626 m/sec^2.
(51.744 - 10.878) = net 40.866N. accelerating force on 5kg.
Acceleration of 5kg. block = (40.866/5) = 8.1732m/sec^2.