Question

honors physics small group review problems
instructions: work on each problem, one at a time, as a group at your white - board station. show all of your work on the board and take turns switching off who writes the problem out. i will give you a check when the problem is completed correctly.

1. an eagle flies in a straight line at a constant speed.

a. is the eagle in equilibrium?
b. what must the net force on the eagle be? (zero or non - zero)

1. a car makes a u turn at a constant speed.

a. is the car in equilibrium?
b. what must the net force on the car be? (zero or non - zero)

1. a shoe has a mass of 0.3 kg and rests on the floor.

a. draw a fbd for the shoe
b. write an equilibrium equation for the shoe
c. calculate the magnitude of the normal force on the shoe from the floor
use the graph below to answer questions 4, 5, and 6
velocity vs. time

1. what is the objects velocity at t = 3 seconds?
2. what is the objects acceleration?
3. calculate the objects displacement from t = 0 seconds to t = 6 seconds

the graph shown below is a position versus time graph. use it to answer questions 7 and 8

1. rank the speed of each of the objects shown in the diagram a, b, c, d, e, and f
2. describe in words what object a, c, and e are doing

Explanation:

1. Problem 1

a.

An object is in equilibrium when it has zero acceleration. Since the eagle is flying in a straight - line at a constant speed, its acceleration \(a = 0\). According to Newton's second law \(F_{net}=ma\), when \(a = 0\), the net force on the eagle is zero, so it is in equilibrium.

b.

From Newton's second law \(F_{net}=ma\), with \(a = 0\) (constant speed in a straight - line motion), \(F_{net}=0\).

2. Problem 2

a.

When a car makes a U - turn at a constant speed, its direction of motion is changing. Since velocity is a vector quantity (has both magnitude and direction), a change in direction means a change in velocity, and thus the car has a non - zero acceleration. So the car is not in equilibrium.

b.

Since the car has a non - zero acceleration (due to the change in direction) during the U - turn, from Newton's second law \(F_{net}=ma\), the net force on the car \(F_{net}
eq0\).

3. Problem 3

a.

A free - body diagram (FBD) for the shoe resting on the floor has two forces: the force of gravity \(F_g\) acting downwards and the normal force \(N\) acting upwards.

b.

In the vertical direction, since the shoe is at rest (in equilibrium), the sum of the forces in the y - direction \(\sum F_y=N - F_g=0\), where \(F_g = mg\).

c.

Given \(m = 0.3\ kg\) and \(g=9.8\ m/s^2\), \(F_g=mg=0.3\times9.8 = 2.94\ N\). From \(\sum F_y=N - F_g = 0\), we get \(N=F_g=2.94\ N\).

4. Problem 4

From the velocity - time graph, at \(t = 3\ s\), we can read the velocity value. The velocity \(v\) at \(t = 3\ s\) is \(5\ m/s\).

5. Problem 5

The acceleration \(a\) of an object on a velocity - time graph is given by the slope of the graph. The slope \(m=\frac{\Delta v}{\Delta t}\). The initial velocity \(v_0 = 2\ m/s\) at \(t = 0\ s\) and the final velocity \(v = 7\ m/s\) at \(t = 6\ s\). So \(a=\frac{v - v_0}{t}=\frac{7 - 2}{6}=\frac{5}{6}\approx0.83\ m/s^2\).

6. Problem 6

The displacement \(\Delta x\) of an object on a velocity - time graph is given by the area under the graph. The area under the graph from \(t = 0\ s\) to \(t = 6\ s\) is the area of a trapezoid. The formula for the area of a trapezoid is \(A=\frac{(a + b)h}{2}\), where \(a\) and \(b\) are the parallel sides and \(h\) is the height. Here, \(a = 2\ m/s\), \(b = 7\ m/s\) and \(h = 6\ s\). So \(\Delta x=\frac{(2 + 7)\times6}{2}=27\ m\).

7. Problem 7

On a position - time graph, the speed is given by the magnitude of the slope of the graph.
Object A: slope \(m_A=0\) (horizontal line, speed \(v_A = 0\))
Object B: positive slope
Object C: positive slope, less steep than B
Object D: positive slope, steeper than B
Object E: negative slope
Object F: positive slope, less steep than C
Ranking the speeds from greatest to least: \(D>B>F>C>E>A\)

8. Problem 8

Object A: Since the position does not change with time, it is at rest.
Object C: It is moving with a constant positive velocity (positive slope of the position - time graph).
Object E: It is moving with a constant negative velocity (negative slope of the position - time graph), moving in the negative direction.

Answer:

1.
a. Yes
b. Zero
2.
a. No
b. Non - zero
3.
a. FBD: Gravity downwards, normal force upwards
b. \(\sum F_y=N - mg=0\)
c. \(2.94\ N\)

1. \(5\ m/s\)
2. \(\frac{5}{6}\ m/s^2\approx0.83\ m/s^2\)
3. \(27\ m\)
4. \(D>B>F>C>E>A\)

8.
Object A: At rest
Object C: Moving with constant positive velocity
Object E: Moving with constant negative velocity