course Phy 121
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vector A, with x and y components 9.2 m/s and 9.2 m/s, and
vector B, whose x and y components are 5 m/s and .7999998 m/s?
What are the magnitude and angle of the resultant vector?
sum x components: 14.2 m/s
sum y components: 9.999 m/s
Angle: arctan y/x = arctan 9.99 m/s / 14.2 m/s = 35 degrees
Magnitude: a^2+b^2= c^2, so sqrt(14.2^2 + 9.99^2) = 17.4
Problem Number 2
A velocity vector has magnitude 9 and is directed at an angle with the positive x axis of 16 degrees. What are the velocities in the x and y directions of the matching motions in those directions?
Ax = A cos(theta) = 9 cos(16) = 8.7 m/s
Ay = A sin(theta) = 9 sin(16) = 2.5 m/s
Problem Number 3
An object oringinally moving at a constant speed is acted upon for a specified time bya constant force of 90 Newtons.
At the end of the specified time the force is removed and the object procedes at a new constant velocity.
If the oject traveled a distance of 120 meters while under the influence of the force, and if 7128 Joules if kinetic energy are dissipated to friction, then by how much will the kinetic energy of the object increase>
Joules: 90N * 120m = 10800 Joules or Newton meters
If 7128 Joules are dissipated: 108000 - 7128 = 3672 Joules of increased kinetic energy.
Problem Number 4
An object is being pushed by a force of 3 Newtons, directed parallel to its direction of motion.
We assume that the frictional and other dissipative forces are negligible, to the object gains in energy all the work done by the force.
How far will the object have to be pushed in order to gain 17 Joules of energy?
3 Newtons (parallel to its direction of motion) so, 17 Joules / 3 Newtons = 5.66 meters
Problem Number 5
If I walk 5.99 miles to the East, then 5.08 miles to the North, I will have walked 11.07 miles.
If you walk in a straight line from my starting point to my final position, what angle will your path make with East, and how far will you walk?
sqrt(5.99^2 + 5.08^2) = 7.85 mi
arctan 5.08/5.99 = 40.3 degrees
Problem Number 6
A mass of 86 kilograms is moving at 20 meters/second. Find its kinetic energy, its kinetic energy if its speed doubled, and the factor by which kinetic energy increases when speed is doubled.
KE = .5mv^2 = .5(86kg)(20 m/s) = 17,200 kg m/s
KE = .5(86 kg)(40 m/s) = 68,800 kg m/s
68,800 / 17,200 = 4 so the kinetic energy increases by a factor of 4.
Problem Number 7
An object of mass 4 kg is acted upon by an unknown force F for 'dt = .07 seconds. Its velocity is observed to change during this time from 7 m/s to 6.405 m/s.
Use the Impulse-Momentum Theorem to determine the average force exerted on the object.
Verify your results using you knowledge of uniformly accelerated motion.
m = p/v so p = m*v p = 4 kg * .595 m/s = 2.38 kg m/s
Good, but you would want to express this in terms of `dv and `dp, change in velocity and change in momentum:
m = `dp/`dv so `dp = m * `dv; `dp = 4 kg * .595 m/s = 2.38 kg m/s.
This puts the calculation into a format consistent with the rest of your solution.
Then: F = dp / dt = 2.38 kg m/s / .07 s = 34 Newtons/sec
Force = m*a and a = change in velocity / time = .595 m/s / .07s = 8.5 m/s/s so that F = 4 kg * 8.6 m/s/s = 34 Newtons/sec
Problem Number 8
What is the change in the velocity of an object of mass 8 kg which experiences a variable net force F(t) for .6 seconds, if the force has the following characteristics :
The force vs. clock time graph increases linearly from 0 Newtons at the beginning of the .6 second interval to 129.6 Newtons, then decreases linearly back to 0 at the end of the time interval.
F*dt = dp so that 129.6 N * .6 sec = 77.76 Newton sec
v = p/m = 77.76 Newton sec/ 8 kg = 9.72 kg N sec
`dp = F_ave * `dt, and since the force changes linearly from 0 N to 129.6 N on one half of the interval, and from 129.6 N to 0 N on the other half, the average force for each half of the interval is (129.6 N + 0 N) / 2 = 64.8 N.
The average on both halves is the same, so the average on the entire interval is the same.
You therefore use 64.8 N as the average force, not 129.6 N.
Otherwise your solution is correct.