Solving Difficult Physics Problems

Solving Difficult Physics Problems

Here, we are going to consider the same relatively simple physics problem in a variety of different ways with increasing complexity. We’ll see the problem can be solved from multiple perspectives and we’ll see how good are low order approximations really are.

Given the setup on the left, we want this problem with increasing complexity. First, we will neglect friction and treat the pulley as massless. Then we will add friction. Then we add the moment of inertia of the pulley. We consider both the .5 kg mass and the 1.4 kg mass to be at rest before we begin the scenario. The 1.4 kg block then free falls 88 cm (we can neglect air resistance over this short journey even in our most sophisticated approximation).

0. What makes a problem difficult and how do we solve it?

There are several different factors which contribute to making a problem difficult, but the principal one is that planning and patience are required to solve the problem. Difficult problems often require subdividing a large problem into smaller pieces which can be contended with using known information.

  • First, gather up all known information from the problem.
  • Next, gather relevant resources to solve problem and choose an approach.
  • Then look for ways to break the problem into smaller, more manageable chunks.
  • If the problem still seems intractable, look for places where I might make simplifying assumptions in the problem that lead to a related, simpler, solvable problem.

Difficult problems require multiple steps, but you should not be daunted by them. The real world is messy and full of difficult problems that require nuance and attention to detail. Solving difficult problems in an academic setting helps to prepare you to solve difficult problems in the real world. It is good to be mentally frustrated and challenged. It helps you grow as a problem solver.

1. The total length of the inclined plane is 3m (along the hypotenuse). Determine how long (in seconds to two decimal places) the .5 kg block spends sliding along the inclined plane neglecting friction and treating the pulley as massless. When the 1.4 kg block hits the floor, the rope disconnects from the block. Determine which side of the inclined plane the block would exit from. Assume the experiment is done on Earth.

First, we have to set up an initial systems of equations where we realize that the relationship between those systems is dictated by the equality of the tension in the rope holding the two masses together. We will apply a net force equation and then Newton’s second law to each mass. But notice this is a HARD problem because we need to divide the path of the second block into two different paths- the first 88 cm of the .5 kg block’s path is under constant acceleration from the 1.4 kg block falling. The next bit of the path is under constant acceleration in the opposite direction from gravity.

$F_{net\text{ .5 kg}} = F_{parallel} – F_T$

$F_{net\text{ .5 kg}} = mgsin(34^0) – F_T$

$.5a = .5(9.8)sin(34^0) – F_T$

In the next step assigning the proper sign to tension is crucial. We must realize that generally we will assign a negative sign to downwards motion and that tension here opposes that motion, so we have:

$F_{net\text{ 1.4 kg}} = F_T- F_g$

Moreover, it was important that the sign of $F_T$ was different in the two differing equations because it is acting on the masses in opposite directions.

$F_{net\text{ 1.4 kg}} = F_T – mg$

$1.4a = F_T- 1.4(9.8)$

$1.4(9.8) + 1.4a = F_T$

$.5a = .5(9.8)sin(34^0) – 1.4(9.8) – 1.4a$

$.5a = 2.74 – 13.72 – 1.4a$

$1.9a = -10.98 \rightarrow a = -5.78 \text{ }\frac{m}{s^2}$

This acceleration means the mass on the inclined plane is heading rightwards (at least for the first 88 or so cm!). Now, we need to resort to kinematics.

$v_f^2 = v_i^2 + 2as$

$v_f = -\sqrt{2(5.78)(.88)} =-3.19 \text{ }\frac{m}{s}$

But we’ve only covered 88 cm and 130 cm remain. Is the box moving fast enough to escape the ramp rightwards? Now, we have only the parallel force acting on the box.

$F_{parallel} = mgsin(34^0) = ma \rightarrow a = 5.48 \text{ }\frac{m}{s^2}$.

Solving the next kinematics equation for $s$ when $v_f = 0$ tells us how long the box needs to start sliding the other way.

$0 = -(3.19^2) + 2(5.48)s \rightarrow s = .93 m$

Now, we know the box does not make it all the way up the ramp, but starts to slide down the ramp again under the influence of gravity. This means it now slides backwards 93 cm + 88 cm + 82 cm = 263 cm. And here, we have:

$v_f^2 = v_i^2 + 2as$

$v_f = \sqrt{2(5.48)(2.63)} = 5.37 \text{ }\frac{m}{s}$

Now, we’re asked for the total time and there are 2 phases of movement here, each under constant acceleration. In the first phase, we got from 0 $\frac{m}{s}$ to -3.19 $\frac{m}{s}$ under a constant acceleration of -5.78 $\frac{m}{s^2}$. In the second phase, we go from -3.19 $\frac{m}{s}$ to 5.37 $\frac{m}{s}$ under a constant acceleration of 5.48 $\frac{m}{s^2}$. So we utilize

$v_f = v_i + at$

$-3.19 = 0 + -5.78t_1 \rightarrow t_1 = .552 s$

$5.48 = -3.19 + -5.37t_2 \rightarrow t_2 = 1.615 s$

So, the total time spent sliding on the inclined plane is $t_1 + t_2 = 2.17 s$ and it exits the inclined plane to the left.

2. The total length of the inclined plane is 3m (along the hypotenuse). Determine how long (in seconds to two decimal places) the .5 kg block spends sliding along the inclined plane considering friction and using a coefficient of friction of $\mu = .15$ and treating the pulley as massless. When the 1.4 kg block hits the floor, the rope disconnects from the block. Determine which side of the inclined plane the block would exit from. Assume the experiment is done on Earth.

In theory, we have much the work already done for us with the last problem.

$F_{net\text{ .5 kg}} = F_{parallel} + F_f – F_T$

$F_{net\text{ .5 kg}} = mgsin(\theta) + \mu mgcos(\theta) – F_T$

$.5a = .5(9.8)sin(34^0) + .15(.5)(9.8)cos(34^0) – F_T$

$F_{net\text{ 1.4 kg}} = F_T- F_g$

$F_{net\text{ 1.4 kg}} = F_T – mg$

$1.4a = F_T- 1.4(9.8)$

$1.4(9.8) + 1.4a = F_T$

$.5a = .5(9.8)sin(34^0) + .15(.5)(9.8)cos(34^0) – 1.4(9.8) – 1.4a$

$.5a = 2.74 + .61 – 13.72 – 1.4a$

$1.9a = -10.37 \rightarrow a = -5.46 \text{ }\frac{m}{s^2}$

Now, we need to resort to kinematics.

$v_f^2 = v_i^2 + 2as$

$v_f = -\sqrt{2(5.48)(.88)} =-3.11 \text{ }\frac{m}{s}$

But we’ve only covered 88 cm and 130 cm remain. Is the box moving fast enough to escape the ramp rightwards? Now, we have only the parallel force and the friction force acting on the box in the same direction.

$F_{parallel} = mgsin(34^0) + \mu mgcos(34^0) = ma \rightarrow a = 6.70 \text{ }\frac{m}{s^2}$.

Solving the next kinematics equation for $s$ when $v_f = 0$ tells us how long the box needs to start sliding the other way.

$0 = -(3.11^2) + 2(6.70)s \rightarrow s = .72 m$

Now, we know the box does not make it all the way up the ramp, but starts to slide down the ramp again under the influence of gravity, but with friction force opposing its motion. This means it now slides backwards 72 cm + 88 cm + 82 cm = 242 cm. And here, we have:

$F_{parallel} – F_f = mgsin(/theta) – \mu mgcos(\theta) = ma \rightarrow g(sin(34^0 – .15*cos(34^0) = a = 4.26 \text{ }\frac{m}{s}$

$v_f^2 = v_i^2 + 2as$

$v_f = \sqrt{2(4.26)(2.42)} = 4.54 \text{ }\frac{m}{s}$

Now, we’re asked for the total time and there are now 3 distinct phases of movement here due to the opposition of movement caused by friction, each under constant acceleration.

$v_f = v_i + at$

Phase 1: $-3.11 = 0 + -5.48t_1 \rightarrow t_1 = .568 s$

Phase 2: $0 = -3.11 + 6.70t_2 \rightarrow t_2 = .464 s$

Phase 3: $4.54 = 0 + 4.26t_3 \rightarrow t_2 = 1.066 s$

So, the total time spent sliding on the inclined plane is $t_1 + t_2 = 2.09 s$ and it exits the inclined plane to the left.

3. The total length of the inclined plane is 3m (along the hypotenuse). Determine how long (in seconds to two decimal places) the .5 kg block spends sliding along the inclined plane considering friction and using a coefficient of friction of $\mu = .15$ and the pulley is a disk with radius 4 cm and mass 50 g. When the 1.4 kg block hits the floor, the rope disconnects from the block. The rope will slide freely across the pulley if the masses become detached. Determine which side of the inclined plane the block would exit from. Assume the experiment is done on Earth.

Now, we ask ourselves what the introduction of the pulley means. We can no longer assume the tension on one side of the pulley is equal to the tension on the other side. Some of the force from the falling block is used to rotate the pulley. We must add an equation for torque.

Equation 1

$\tau = I\alpha$

$\tau = \frac{1}{2}m_{pulley}r^2\frac{a}{r}$

$T_1r – T_2r = \frac{1}{2}m_pr^2\frac{a}{r}$

$T_1 – T_2 = \frac{1}{2}m_pa$

Equation 2

$F = m_{incline}a$

$F = m_ia$

$T_1 – F_{parallel} – F_f = m_{i}a$

$T_1 – m_i gsin(\theta) – \mu m_i gcos(\theta) = m_ia$

Equation 3

$F = m_{hanging}a$

$m_hg – T_2 = m_ha$

Now, I will add equations 2 and 3 together to obtain equation 4 given below

$T_1 – m_i gsin(\theta) – \mu m_i gcos(\theta) + m_hg – T_2 = m_ia + m_ha$

$T_1 – T_2 – m_i gsin(\theta) – \mu m_i gcos(\theta) + m_hg = m_ia + m_ha$

And here, I substitute into equation 4 with equation 1 for the term $T_1 – T_2$.

$\frac{1}{2}m_pa – m_i gsin(\theta) – \mu m_i gcos(\theta) + m_hg= m_ia + m_ha$

$- m_i gsin(\theta) – \mu m_i gcos(\theta) + m_hg = m_ia + m_ha – \frac{1}{2}m_pa$

$- m_i gsin(\theta) – \mu m_i gcos(\theta) + m_hg= \left( m_i + m_h – \frac{1}{2}m_p\right)a$

$\frac{- m_i gsin(\theta) – \mu m_i gcos(\theta) + m_hg}{ m_i + m_h – \frac{1}{2}m_p}= a$

$\frac{- (.5)(9.8)sin(34^0) – (.15)(.5)(9.8)cos(34^0) + 1.4(0.8)}{.5+ 1.4 – \frac{1}{2}(.05)}= -1.19 \text{ }\frac{m}{s^2}$

Then we apply the same kinematics equation as before to obtain:

$v_f^2 = v_i^2 + 2as$

$v_f = -\sqrt{ 2(1.19)(.88)} = -1.45 \text{ }\frac{m}{s}$

Phase 1 sliding up ramp time:

$v_f = v_i + at_1$

$-1.45 = 0 + -1.19t_1$

$t_1 = 1.219 \text{ s}$

Now, we consider phase 2 of the sliding where the block continues to slide before coming to rest and falling back down the inclined plane. Here, $F_{parallel}$ is identical to the previous problem indicating that our acceleration should be the same as well.

$a = 6.70 \text{ }\frac{m}{s^2}$

$v_f = v_i + at_2$

$0 = -1.45 + 6.7t_2$

$t_2 = .216 \text{ s}$

$v_f^2 = v_i^2 + 2as$

$0 = 1.45^2 + 2(6.7)s$

$s = 16 \text{ cm to the right}$


Finally, we consider phase 3. We know our block is 16 cm + 88 cm + 72 cm from the top bottom of the inclined plane and is acted on by the same forces as in problem 2.

Again, we have

$a = 4.26 \text{ }\frac{m}{s}$

$v_f^2 = v_i^2 + 2as$

$v_f = \sqrt{2*1.76*4.26} = 3.87 \text{ }\frac{m}{s}$

$v_f = v_i + at_3$

$3.87 = 0 + 4.26t_3$

$t_3 = .908 \text{ s}$

Thus the total time is .126 + .908 + 1.219 s = 2.25 3