1.25 m A 20.0-g object is placed against the free end of a spring (with spring c

Question

1.25 m A 20.0-g object is placed against the free end of a spring (with spring constant k equal to 25.0 N/m) that is compressed 10.0 cm. Once released, the object slides 1.25 m across the tabletop and eventually lands 1.21 m from the edge of the table on the floor, as shown in the figure Calculate the coefficient of friction between the table and the object. The sliding distance includes the compression of the spring and the tabletop is 1.00 m above the floor level 100001 1.00 m Number 1.21 m here is a hint available

Explanation / Answer

Solution: Upward and rightward directions are to be taken positive and downward and leftward direction are taken to be negative.

Mass of the pbject m = 20.0 g = 0.02 kg

The vertical height fallen by the object from the edge of the table h = -1m

Horizontal distance travelled by the while falling from the table x = 0.61 m

Suppose the object leaves the edge of the table with the speed of v.

Since it makes angle of = 0o while it leaves the table edge,

vertical component of speed vy = v*sin

horizontal component of speed vx = vcos

For the vertical motion, we have,

h = (v*sin)*t + (1/2)*gt2

-1m = (v*sin0o) + (1/2)*(-9.81m/s2)*t2

t2 = 0.20387

t = 0.4515 s

Thus the object takes 0.4515 s to fall vertical distance of 1m.

Same time is taken by it to cover horizontal distance till hits the floor.

For horizontal motion we have,

vx = d/t

v*cos0o = 0.61m / 0.4515s

v = 1.35098 m/s

Thus the object leaves the edge of the table with the speed of 1.35098 m/s

The kinetic energy of the object while it is leaving the edge of the table is

Kf = (1/2)*m*v2

Kf = (1/2)*(0.02kg)*(1.35098m/s)2

Kf = 0.0182515 J

Now we consider the initial situation.

The spring constant of the spring k = 25.0 N/m

When the spring is compressed d = 100cm = 0.1 m, the elastic potential energy is stored inside it.

U = (1/2)*k*d2

U = (1/2)*(25.0Nm)*((0.1m)2

U = 0.125 J.

When the spring is released, all of its elastic potential energy is transferred into the object’s kinetic energy Ki and thus object gains some initial speed.

U = Ki                               thus,

Ki = 0.125 J

We notice that the initial kinetic energy of the object greater than its final kinetic energy when it leaves the table. Part of its initial kinetic energy is spent while doing the work against the frictional force f.

The work done W against the frictional forces is is simply the difference between the initial and final kinetic energy of the object

W = Ki - Kf

W = 0.125 J - 0.0182515 J

W = 0.1067485 J

Now we have, work done by frictional forces W = f*l where l = 1.25 m (object slides on table)

but the frictional force is f = *mg

W = f*l

W = (*m*g)*(l)

0.1067485 J = (*0.02kg*9.81m/s2)*(1.25m)                    

= 0.4353

thus the coefficient of friction is 0.4353

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