PLEASE ANSWER ENTIRE QUESTION!!! A magnet of mass 0.9 kg is suspended from the c

Question

PLEASE ANSWER ENTIRE QUESTION!!! A magnet of mass 0.9 kg is suspended from the ceiling by a cord of length L = 2.2 meters. A large magnet is somewhere off to the right, pulling on the small hanging magnet with a constant force of Fm = 2.5 Newtons. Assume that gravity is g = 9.8 m/s2. (a) At what angle with respect to the vertical does the magnet hang? (b) What is the magnitude of the tension in the cord by which the magnet hangs? Suppose now that the cord is cut, and the magnet begins to fall toward the ground while simultaneously being pulled to the right with the same constant force of Fm = 2.5 Newtons. The ceiling is 7.2 meters above the floor. Ignore the effects of air resistance. (c) How long will it take the magnet to reach the floor? (d) How far will the magnet have traveled horizontally from its start point when it hits the floor? (e) When the magnet hits the floor, it continues being pulled to the right by the same magnetic force as before. The coefficient of friction between the magnet and the floor is 0.22. What will the magnet’s acceleration be as it slides to the right along the floor?

Explanation / Answer

(a) Say the angle made by the cord is with the vertical.
Now Say the tension in the string is T.
The gravity force on the magnet will be downward and it will be Mg .

where M is the mass of the magnet.
Now resolve the Tension T into its components along vertical and horizontal

TCos = Mg       .....(1)
TSin = Fm = 2.5             ..... (2)

therefore dividing 2 by 1
tan = 2.5/Mg

= tan^-1 [ 2.5 / (0.9*9.8)]
= 15.83 degree

b) The magnitude of tension in the chord is

T Sin = Fm

T = 2.5/ sin 15.83

T = 9.17 N

c) h = 7.2 - 2.2 cos 15.83

h = 5.083 m

t = sqrt(2h/g)

t = sqrt(2*5.083 / 10)

t = 1.008 s

d) a = 2.5/0.9

a = 2.78

Horizontal Distance = 0.5*a*t^2

                           = 0.5*2.78*1.008^2

                          = 1.41 m

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