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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.
Solution:
ma | = | T - mg | |
a | = | - g | |
= | - 9.8 m/s ^{2} = 3.8 m/s ^{2} | (5) |
v | = | v_{0} + 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.
Solution:
v^{ 2} | = | v_{0}^{2} + 2ax | |
a | = | = = - 0.6 m/s ^{2} | (7) |
f_{k} = ma = .5(- 0.6) = - 0.3 N. | (8) |
- mg | = | ma | |
= | - a/g | ||
= | 0.061. | (9) |
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}.
Solution:
From the 2nd Law:
F_{s} - mg | = | ma | |
F_{s} | = | 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.
Solution:
w_{x} | = | - mgsin | |
w_{y} | = | - mgcos . |
x | : - mgsin + f_{k} = - 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.
Solution:
We apply the 2nd law separately to each box.
For the 10 kg box:
y direction:
N_{2} - m_{2}g | = | ||
N_{2} | = | m_{2}g, |
T - f_{k 2} | = | m_{2}a | |
T - N_{2} | = | m_{2}a | |
T - m_{2}g | = | m_{2}a. | (13) |
N_{1} = m_{1}gcos , |
m_{1}gsin - T - f_{k 1} | = | m_{1}a | |
m_{1}gsin - T - N_{1} | = | m_{1}a | |
m_{1}gsin - T - m_{1}gcos | = | m_{1}a. | (14) |
We have a system of two equations and two unknowns: a and T . We can
solve
as follows.
m_{1}a | = | m_{1}gsin - m_{1}gcos - m_{2}a - m_{2}g | |
m_{1}a + m_{2}a | = | m_{1}gsin - m_{2}g - m_{1}gcos | (15) |
a | = | [m_{1}gsin - m_{2}g - m_{1}gcos ]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 | = | m_{2}(a + g) | |
= | 10(9.8(.1) + 1.1) = 20.8 N. | (17) |
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