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A box of mass 5.0 kg is pulled vertically upwards by a force of 68 N applied to a rope attached to the box. Find a) the acceleration of the box and b) the vertical velocity of the box after 2 seconds.
|ma||=||T - mg|
|=||- 9.8 m/s 2 = 3.8 m/s 2||(5)|
|v||=||v0 + at|
|=||0 + 3.8(2) = 7.6 m/s.||(6)|
A hockey puck of mass .5 kg travelling at 10 m/s slows to 2.0 m/s over a distance of 80 m. Find a) the frictional force acting on the puck and b) the coefficient of kinetic friction between the puck and the surface.
|v 2||=||v02 + 2ax|
|a||=||= = - 0.6 m/s 2||(7)|
|fk = ma = .5(- 0.6) = - 0.3 N.||(8)|
A student of mass 50 kg tests Newton's laws by standing on a bathroom scale in an elevator. Assume that the scale reads in newtons. Find the scale reading when the elevator is a) accelerating upward at .5 m/s 2, b) going up at a constant speed of 3.0 m/s and c) going up but decelerating at 1.0 m/s 2.
From the 2nd Law:
|Fs - mg||=||ma|
|Fs||=||m(g + a).||(10)|
A wooden plank is raised at one end to an angle of 30 o . A 2.0 kg box is placed on the incline 1.0 m from the lower end and given a slight tap to overcome static friction. The coefficient of kinetic friction between the box and the plank is = 0.20 . Find a) the rate of acceleration of the box and b) the speed of the box at the bottom. Assume that the initial speed of the box is zero.
|wy||=||- mgcos .|
|x||: - mgsin + fk = - ma|
|y||: N - mgcos = 0.||(11)|
|ma||=||- (mgcos ) + mgsin|
|a||=||g(sin - cos )|
|=||9.8(sin 30 - 0.2cos 30) = 3.20 m/s 2||(12)|
|v = = 2.53 m/s .|
A 10 kg box is attached to a 7 kg box which rests on a 30 o incline. The coefficient of kinetic friction between each box and the surface is = .1 . Find a) the rate of acceleration of the system and b) the tension in the rope.
We apply the 2nd law separately to each box.
For the 10 kg box:
|N2 - m2g||=|
|T - fk 2||=||m2a|
|T - N2||=||m2a|
|T - m2g||=||m2a.||(13)|
|N1 = m1gcos ,|
|m1gsin - T - fk 1||=||m1a|
|m1gsin - T - N1||=||m1a|
|m1gsin - T - m1gcos||=||m1a.||(14)|
We have a system of two equations and two unknowns: a and T . We can
|m1a||=||m1gsin - m1gcos - m2a - m2g|
|m1a + m2a||=||m1gsin - m2g - m1gcos||(15)|
|a||=||[m1gsin - m2g - m1gcos ]g]|
|=||[7(9.8)sin 30 - (0.1)(10)(9.8) - (0.1)(7)(9.8)cos 30)] = 1.1 m/s 2||(16)|
|T||=||m2(a + g)|
|=||10(9.8(.1) + 1.1) = 20.8 N.||(17)|