A 7.5 kg block is resting on a horizontal surface, and is connected to a 5.5 kg mass suspended from a string..

Question

passed over a frictionless pulley.
Find the acceleration of the system and the tension in the string if the surface is:
(a)
smooth, and
(b)
the coefficient of friction is 0.190

Answer 1

The resulting force on this system (total mass M = 5,5 kg + 7,5 kg = 13 kg) is determined by friction "Fr" of the 7.5 kg mass and gravitational pull "Fg" on the 5.5 kg mass.
So in the following, use these values for Fg and Fr in your situation:
M = 13 kg
Fg is 9,81 N/kg * 5,5 kg = 53,96 N
Fr is 9,81 N/kg * 7,5 kg * 0,190 = 13,98 N

a) Acceleration a is Fg / M = 53,96 N / 13 kg = 4,15 m/s^2
b) Acceleration is (Fg - Fr) / M = 39,98 N / 13 kg = 3,08 m/s's.

I hope this helps.

Answer 2

As F = ma --------------------->a million we've mass, yet we could desire to be certain acceleration.. For this use equation; d = a million/2at^2 ....... right here t is the time, d is the displacement and a is acceleration 2d = at^2 2d / t^2 = a --------> 2 Now exchange this fee in equation --->a million F = ma F = m(2d/t^2) F = 2md / t^2 Now placed all values F = 2(fifty 5)(32) / (5.3)^2 F = 3520 / 28.09 F = one hundred twenty five.3115 Newton

Answer 3

http://en.wikipedia.org/wiki/Coefficient_of_friction