Forces & Kinematics: Videos & Practice Problems

In solving one-dimensional motion problems involving forces, utilize the equation $F=ma$ alongside the equations of uniformly accelerated motion (UAM). Key variables include initial velocity, final velocity, displacement, time, and acceleration. For example, a 20 kg block accelerating from rest to 30 m/s in 6 seconds requires calculating acceleration as the link between force and motion. This results in an applied force of 100 N, demonstrating the integration of force and motion concepts effectively.

Solving Motion Problems with Forces

In solving one-dimensional motion problems that involve forces, it's essential to integrate the concepts of motion and force. The five key motion variables to remember are initial velocity (\(v_{\text{initial}}\)), final velocity (\(v_{\text{final}}\)), displacement (\(\Delta x\)), time (\(\Delta t\)), and acceleration (\(a\)). When you know three of these variables, you can apply one of the four equations of uniformly accelerated motion (UAM) to find the unknown variable.

On the forces side, the primary variables are force (\(F\)), mass (\(m\)), and acceleration (\(a\)), governed by the equation \(F = ma\). To solve a forces problem, you need to know two of these three variables. In scenarios where both motion and forces are involved, the acceleration variable serves as the common link between the two sets of equations.

For example, consider a 20 kg block on a frictionless surface being pushed and accelerating from rest to a final velocity of 30 m/s over a time period of 6 seconds. To find the applied force, you first draw a free body diagram to identify the forces acting on the block, which include the weight force and the applied force, while noting that there is no friction due to the frictionless surface.

Next, you apply the equation \(F = ma\) to find the applied force. However, since the acceleration is unknown, you can use the motion equations to find it. Using the equation \(v_{\text{final}} = v_{\text{initial}} + at\), you can substitute the known values: \(30 = 0 + a \cdot 6\). Solving for \(a\) gives you \(a = \frac{30}{6} = 5 \, \text{m/s}^2\).

Now that you have the acceleration, you can substitute it back into the force equation: \(F_a = m \cdot a = 20 \cdot 5 = 100 \, \text{N}\). Thus, the magnitude of the applied force pushing the block is 100 newtons. This example illustrates how to effectively combine the principles of forces and motion to solve complex problems.

An 800-kg car is traveling along a horizontal road directly towards a cliff. The driver notices and brakes, resulting in a 5,000-N net force slowing the car down. If the car's initial speed was 20 m/s and the car stops just before going over the cliff, how far away was the car from the cliff when the driver hit the brakes?

• A. 1250m
• B. 0.03m
• C. 64m
• D. 32m

Answer: (D)

Liftoff!

In this scenario, we analyze a 1,000-kilogram rocket that is accelerating upwards due to thrust. The forces acting on the rocket include the weight force (force of gravity), thrust, and air resistance. The weight force, denoted as \( w \), is calculated as \( w = mg \), where \( g \) is the acceleration due to gravity, typically approximated as \( 10 \, \text{m/s}^2 \). Thus, for our rocket, \( w = 1,000 \, \text{kg} \times 10 \, \text{m/s}^2 = 10,000 \, \text{N} \).

The thrust force, \( F_t \), is given as \( 25,000 \, \text{N} \), while the air resistance, \( F_{\text{air}} \), is \( 5,000 \, \text{N} \). To find the net force acting on the rocket, we consider the upward direction as positive. Therefore, the net force \( F_{\text{net}} \) can be expressed as:

\( F_{\text{net}} = F_t - F_{\text{air}} - w \)

Substituting the known values, we have:

\( F_{\text{net}} = 25,000 \, \text{N} - 5,000 \, \text{N} - 10,000 \, \text{N} = 10,000 \, \text{N} \)

Using Newton's second law, \( F = ma \), we can solve for acceleration \( a \):

\( 10,000 \, \text{N} = 1,000 \, \text{kg} \times a \)

From this, we find:

\( a = \frac{10,000 \, \text{N}}{1,000 \, \text{kg}} = 10 \, \text{m/s}^2 \)

Now that we have the acceleration, we can use kinematic equations to find the final velocity after 20 seconds. The initial velocity \( v_0 \) is \( 0 \, \text{m/s} \), and the time \( t \) is \( 20 \, \text{s} \). We can use the equation:

\( v_f = v_0 + at \)

Substituting the known values:

\( v_f = 0 + (10 \, \text{m/s}^2)(20 \, \text{s}) = 200 \, \text{m/s} \)

Thus, the final velocity of the rocket after 20 seconds is \( 200 \, \text{m/s} \).

Here’s what students ask on this topic:

What are the equations of uniformly accelerated motion (UAM)?

The equations of uniformly accelerated motion (UAM) are essential for solving kinematics problems. They are:

1. $v_f=v_i+at$

2. $v_{f}2=v_{i}2+\; 2a\mathrm{\Delta x}$

3. $\mathrm{\Delta x}=v_{i}t+\; 0.5at2$

4. $\mathrm{\Delta x}=v_{f}t-\; 0.5at2$

5. $\mathrm{\Delta x}=\; 0.5(v_i+v_f)t$

How do you solve a problem involving both forces and motion?

To solve a problem involving both forces and motion, follow these steps:

1. Identify the known variables for both forces (F, m, a) and motion (v_i, v_f, Δx, Δt, a).

2. Draw a free-body diagram to visualize the forces acting on the object.

3. Use $F\; =\; ma$ to relate the forces to the acceleration.

4. If acceleration (a) is unknown, use one of the UAM equations to solve for it.

5. Substitute the acceleration back into $F\; =\; ma$ to find the force.

By combining these steps, you can effectively solve problems that integrate both force and motion concepts.

What is the significance of the acceleration variable in solving combined force and motion problems?

The acceleration variable (a) is crucial in solving combined force and motion problems because it serves as the link between the two sets of equations. In force problems, acceleration is used in $F\; =\; ma$. In motion problems, acceleration appears in the UAM equations. By determining acceleration from motion variables (v_i, v_f, Δx, Δt), you can use it to find the force using $F\; =\; ma$. This interconnectedness allows for a comprehensive approach to solving problems that involve both forces and motion.

How do you draw a free-body diagram for a forces problem?

To draw a free-body diagram for a forces problem, follow these steps:

1. Represent the object as a simple shape (e.g., a box).

2. Draw arrows to represent all forces acting on the object. Label each force (e.g., F_a for applied force, F_g for gravitational force).

3. Include the weight force (F_g = mg) acting downward.

4. Add normal force (F_N) acting perpendicular to the surface.

5. Include any applied or tension forces.

6. If applicable, add frictional forces.

This diagram helps visualize the forces and is essential for applying $F\; =\; ma$ to solve the problem.

What is the relationship between force, mass, and acceleration?

The relationship between force, mass, and acceleration is described by Newton's Second Law of Motion, which states that the net force (F) acting on an object is equal to the mass (m) of the object multiplied by its acceleration (a). This is expressed as:

$F\; =\; ma$

This equation indicates that for a given mass, the acceleration of an object is directly proportional to the net force applied. Conversely, for a given force, the acceleration is inversely proportional to the mass.