course Phy201 ?????O??????C€???assignment #020020. `query 20
......!!!!!!!!...................................
10:20:48 Explain how we get the components of the resultant of two vectors from the components of the original vectors.
......!!!!!!!!...................................
RESPONSE --> ok confidence assessment:
.................................................
......!!!!!!!!...................................
10:21:06 ** If we add the x components of the original two vectors we get the x component of the resultant. If we add the y components of the original two vectors we get the y component of the resultant. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
10:22:20 Explain how we get the components of a vector from its angle and magnitude.
......!!!!!!!!...................................
RESPONSE --> ok confidence assessment:
.................................................
......!!!!!!!!...................................
10:22:28 ** To get the y component we multiply the magnitude by the sine of the angle of the vector (as measured counterclockwise from the positive x axis). To get the x component we multiply the magnitude by the cosine of the angle of the vector (as measured counterclockwise from the positive x axis). **
......!!!!!!!!...................................
RESPONSE --> ook self critique assessment:
.................................................
??K????????l??y?assignment #020 020. Forces (inclines, friction) Physics II 10-26-2008
......!!!!!!!!...................................
10:16:47 `q001. Note that this assignment contains 3 questions. . A 5 kg block rests on a tabletop. A string runs horizontally from the block over a pulley of negligible mass and with negligible friction at the edge of the table. There is a 2 kg block hanging from the string. If there is no friction between the block in the tabletop, what should be the acceleration of the system after its release?
......!!!!!!!!...................................
RESPONSE --> 9.8m/s^2 confidence assessment:
.................................................
......!!!!!!!!...................................
10:17:02 Gravity exerts a force of 5 kg * 9.8 meters/second = 49 Newtons on the block, but presumably the tabletop is strong enough to support the block and so exerts exactly enough force, 49 Newtons upward, to support the block. The total of this supporting force and the gravitational force is zero. The gravitational force of 2 kg * 9.8 meters/second = 19.6 Newtons is not balanced by any force acting on the two mass system, so we have a system of total mass 7 kg subject to a net force of 19.6 Newtons. The acceleration of this system will therefore be 19.6 Newtons/(7 kg) = 2.8 meters/second ^ 2.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
10:18:45 `q002. Answer the same question as that of the previous problem, except this time take into account friction between the block in the tabletop, which exerts a force opposed to motion which is .10 times as great as the force between the tabletop and the block. Assume that the system slides in the direction in which it is accelerated by gravity.
......!!!!!!!!...................................
RESPONSE --> 3.08m/s^2 confidence assessment:
.................................................
......!!!!!!!!...................................
10:19:03 Again the weight of the object is exactly balance by the upward force of the table on the block. This force has a magnitude of 49 Newtons. Thus friction exerts a force of .10 * 49 Newtons = 4.9 Newtons. This force will act in the direction opposite that of the motion of the system. It will therefore be opposed to the 19.6 Newton force exerted by gravity on the 2 kg object. The net force on the system is therefore 19.6 Newtons -4.9 Newtons = 14.7 Newtons. The system will therefore accelerate at rate a = 14.7 Newtons/(7 kg) = 2.1 meters/second ^ 2.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................