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An object of mass 6.0kg is at rest on a rough horizontal table. A horizontal force of 2.4N acts on the mass to the East. The mass reaches a speed of 1.2 ms-1 in a distance of 2.0m. (a) Calculate the acceleration of the mass. (b) What is the net horizontal force acting on the object? (c) What is the frictional force acting on the object?

May 13, 2018

#1
+2

m = 6 kg

F= 2.4 N

x = xo + vo t + 1/2 a t^2       x0 = 0   vo = velocity original = o   so:

x = 1/2 a t^2        x = 2 (given)       (1)

Now      a = (change in v) / (change in t) = 1.2 m/s / t  (2)   Substitute this into (1)

2 m = 1/2 a t^2

2m = 1/2 (1.2 m/s / t) (t^2)

2 = .6 t        so t = 3.333 sec

Remember   F= ma    so:

F = .6 * a           and    a = 1.2/t    and   t  = 3.333

F=   .6 *  1.2/3.333 = 2.16 N = Net Force (East)

Frictional force is     2.4 - 2.16 N =  .24 N  West       (or  -.24 N East)

May 13, 2018
#2
+1

Thank you very much. I envy you guys and i thnik how and when I will be intelligent like you guys... I started studying late. I am 22yrs. I feel like k*****g myself so I can reincarnate and correct my mistakes on time