+48 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?
+36916 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)
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