Box slips down incline. The box: -40kg -V initial = 0.30m/s -Acc. = 5m/s^2 -Move

Question

Box slips down incline.

The box:
-40kg
-V initial = 0.30m/s
-Acc. = 5m/s^2
-Moves length of 7m down slope

The slope
-35deg with respect to horizontal

(kinetic friction exists betw. Box & slope)

i) When the box gets to the bot. of slope, what is its kin. Enrgy?
ii) With respect to the box, what work is done on it by fric. Force?
iii) with respect to the box, what is magn. of the fric. Force acting on it?
iv) with respect to box and slope, what is µ of kin. Force?
v) with respect to box, what amount of work is done on it by grav. Force?
vi) Calc. the chnge in grav. Potential enrgy of the erth-box systm

Explanation / Answer

   The mass of the box, m = 40 kg    Initial velocity, v = 0.3 m/s    Acceleration, a = 5 m/s^2    The length of the path, s = 7 m    The angle of slope, theeta = 37 degrees    Here the body slides downward, so the acceleration is given by the formula    a = g sin(theeta) - µk gcos(theeta)    µk = [g sin(theeta) - a] /g cos(theeta)    µk = [9.8 sin37 - 5] / cos37    µk = 0.115    (1)The kinetic energy of the box = (1/2) mv^2 = (1/2) 40*(0.3)^2                                                                    = 1.8 J    (2)The workdone by the friction force = fs = µkmgcos(theeta)*s                                                           fs = 0.115*40*9.8*cos37*7 = 252 J    (3)The magnitude of friction force, f = µkmgcos(theeta) = 36 N    (4)   a = g sin(theeta) - µk gcos(theeta)          µk = [g sin(theeta) - a] /g cos(theeta)          µk = [9.8 sin37 - 5] / cos37          µk = 0.115    (5)The workdone by gravity = workdone by friction = 36 N    (6) The vertical height of the block when it is at the top of inclined    plane, h = s*sin(theeta)    The change in the gravitational potential energy = m g h                                                                           = m g s sin(theeta)                                                                           = 40*9.8*7*sin37                                                                           = 1651.4 J          µk = [g sin(theeta) - a] /g cos(theeta)          µk = [9.8 sin37 - 5] / cos37          µk = 0.115    (5)The workdone by gravity = workdone by friction = 36 N    (6) The vertical height of the block when it is at the top of inclined    plane, h = s*sin(theeta)    The change in the gravitational potential energy = m g h                                                                           = m g s sin(theeta)                                                                           = 40*9.8*7*sin37                                                                           = 1651.4 J                                                         

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