A box slides horizontally and then up an incline Assuming th

A box slides horizontally and then up an incline. Assuming the parameters provided, determine the following; Velocity of the box at point B; change in energy due to friction between points A and B; How far up the ramp the package slides before coming to a stop, D;

Solution

A.

mass m = 25/9.8 = 2.551 kg

The velocity at B is,

0.5mvA^2 - ukmgd = 0.5mvB^2

vA^2 - 2u_kgd = v_B^2

here, vA = 15 m/s, uk = 0.45, g = 9.8, d = 14 m,

so, the velocity vB is,

vB = 10.0757 m/s = 10.07 m/s

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B.

From work energy theorem,

W = 0.5mvB^2 - 0.5mvA^2

= 0.5*m[vB^2 - vA^2]

= -157.5 J

Here negative sign indicates the lost of energy. energy is, 157.5 J

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c.

From the law of conservation f energy:

0.5mvB^2 = mgh

  0.5mvB^2 = mg[D*sin35]

0.5vB^2 = g[D*sin35]

0.5*10.07^2 = 9.8*D*sin35

hence, the distance: D = 9.02 m