Question 23 of 26 Map University Physics presented by Sapling Leaming 1.25 m A 2

Question

Question 23 of 26 Map University Physics presented by Sapling Leaming 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 0.61 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 0.61 m + Previous & Give Up & View Solution- Check AnswerNext Ex Hint

Explanation / Answer

Time taken in travelling 1m vertical distance,

h = 1/2gt^2

t = sqrt(2h/g)

t = sqrt(2*1/9.81)

t = 0.45 sec

So the velocity when it leaves the table which is horizontal,

v = d/t

v = 0.61 m / 0.45 s

v = 1.356 m/s

So the kinetic energy when it just leaves the table,

KE = 1/2mv^2

KE = 1/2 x 20 x 10^-3 x 1.356^2

KE = 0.0184 J

The energy stored in spring in the starting,

E_spring = 1/2kx^2 = 0.5 x 25 x 0.1^2 = 0.125 J

So work done by friction force = 0.125 - 0.0184 = 0.1066 J

The magnitude of friction force = _mu*N = mu*mg = mu*20*10^-3*9.81

work done by friction force = mu*20*10^-3*9.81*1.25 = 0.1066

hence mu = 0.1066 / (20 x 10^-3 x 9.81 x 1.25)

mu = 0.435

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