Why Can't I Solve These Seemingly Simple Physics Problems?

[Frustrated]

Simple problems that I can't figure out! HELP!

Hello, first of all, thanks in advance for checking out my thread. below are a short list of some simple problems that should be easy to solve for those of you who frequent this forum. However, I can't seem to figure them out and it is really getting on my nerves. If you have any ideas, i would really appreciate your help. Included with the questions are the numerical answers. As such, I am solely interested in the steps and formulas required for them. THANKS!

Q) A hockey stick gives a hockey puck an initial velocity of 12m/s. If the coefficient of kinetic friction is 0.2, how far will the puck travel before stopping? (ANS = 36m)

Q) A uniform 25 kg board is 12m long and has a 50kg weight attached to one end. How far from the weight will be the center of gravity of this system? (ANS = 2)

Q) A block of mass 4 kg rests on a horizontal surface. The coefficient of sliding kinetic friction between the two is 0.2. A string attached to the block is pulled horizontally by a force with an acceleration of 3m/s2. What is the magnitude of this force? (ANS = 20N)

Q) A CD and spindal together have a moment of inertia of 2x10-4kgm2. They rotate with an angular velocity of 4 rad/sec. What is the net torque that must be applied to bring the system to rest within 2 s? (ANS = 4x10-4 Nm)

Q) A 2.0m wire segment carrying current of 0.5A is oriented at an angle of 30o to a uniform magnetic field of 0.5T. What is the magnetic force of the wire? (cos30 = 0.87, sin30 = 0.5) (ANS = 0.5N)

Q) Given the quantum mechanical model of the hydrogen atom, if the orbital quantum number is 4, then how many diff orbital magnetic quantum numbers are permitted? (ANS = 9)

 

[Pyrrhus]

Q) A hockey stick gives a hockey puck an initial velocity of 12m/s. If the coefficient of kinetic friction is 0.2, how far will the puck travel before stopping? (ANS = 36m)

Use

\Delta E = W_{f}

Q) A uniform 25 kg board is 12m long and has a 50kg weight attached to one end. How far from the weight will be the center of gravity of this system? (ANS = 2)

Use

\vec{r}_{cm} = \frac{\sum_{i=1}^{n} m_{i} \vec{r}_{i}}{\sum_{i=1}^{n} m_{i}}

Q) A block of mass 4 kg rests on a horizontal surface. The coefficient of sliding kinetic friction between the two is 0.2. A string attached to the block is pulled horizontally by a force with an acceleration of 3m/s2. What is the magnitude of this force? (ANS = 20N)

Use

\sum_{i=1}^{n} \vec{F}_{i} = m \vec{a}

Q) A CD and spindal together have a moment of inertia of 2x10-4kgm2. They rotate with an angular velocity of 4 rad/sec. What is the net torque that must be applied to bring the system to rest within 2 s? (ANS = 4x10-4 Nm)

Use

\sum_{i=1}^{n} \vec{\tau}_{i} = I \vec{\alpha}

\sum_{i=1}^{n} \vec{\tau}_{i} = I \frac{d \vec{\omega}}{dt}

\sum_{i=1}^{n} \vec{\tau}_{i} = I \frac{\Delta \vec{\omega}}{\Delta t}

 

[pervect]

You might also find

http://hyperphysics.phy-astr.gsu.edu/hbase/frict2.html

useful, as it has the formulas and explanation of kinetic and static friction. The formulas are simple though, the frictional force is the normal force multiplied by the appropriate coefficient of friction - static, if the object isn't moving, and kinetic, if the object is moving.

The formulas for static and kinetic friction, plus the concept that work = force * distance, and force = mass * acceleration, should take you a long way to solving these problems.

 

[rcgldr]

That last question was a simple one?

 

[rcgldr]

A lot of these constant friction problems can be simplified if you use the fact that the rate of decleration from friction is the coefficient timea the acceleration of gravity (9.8 m/s^2). Since the deceleration is constant, the time is takes to stop is the initial speed divided by the rate of decerleration, the average speed is 1/2 the initial speed, and the distance is the time it takes to stop times the average speed.

For the first question, the coefficient is .2, so the rate of deceleration is:

.2 x 9.8m/s^2 = 1.96 m/s^2

The initial velocity is 12 m / s, so the time it takes to stop is:

12 / 1.96 ~= 6.12 s

Average speed with linear deceleration is 1/2 initial speed or:

12 * 1/2 = 6 m / s

So the distance traveled is average speed times the time it took to stop:

6 m / s time 6.12 s ~ 36.7 m.

If the 1.96 is rounded to 2.0, you get 36m.

 

[Frustrated]

Wow, thanks guys! Much appreciated!