1. you are riding in an elevator holding a spring scale with a 1 - kg mass suspended from it. you see that…

2.

Explanation:

Step1: Recall weight - mass relationship

On Earth, the weight $W = mg$, where $m$ is mass and $g$ is gravitational acceleration ($g_E=9.8\ m/s^2$ on Earth and $g_M=\frac{1}{6}g_E$ on the Moon). For a $10 - kg$ rock on Earth, $W_E=mg_E$. Substituting $m = 10\ kg$ and $g_E = 9.8\ m/s^2$, we get $W_E=10\times9.8=98\ N$. On the Moon, $W_M=mg_M$, and since $g_M=\frac{1}{6}g_E$, $W_M = 10\times\frac{9.8}{6}\approx16.3\ N$. The force holding the rock on Earth is its weight on Earth ($98\ N$) and on the Moon is its weight on the Moon ($16.3\ N$).

3.

Explanation:

Step1: Analyze forces on the mass in the elevator

The normal force $N$ measured by the scale is related to the acceleration of the elevator. The force acting on the $1 - kg$ mass is given by Newton's second - law $F = ma$. The gravitational force on the $1 - kg$ mass is $F_g=mg$ where $m = 1\ kg$ and $g = 9.8\ m/s^2$, so $F_g=9.8\ N$. The normal force $N = 9.3\ N$.

Step2: Apply Newton's second - law

$F_{net}=ma$, and $F_{net}=N - mg$. So $a=\frac{N - mg}{m}$. Substituting $N = 9.3\ N$, $m = 1\ kg$ and $g = 9.8\ m/s^2$, we have $a=\frac{9.3-9.8}{1}=- 0.5\ m/s^2$. The negative sign indicates that the elevator is accelerating downwards. This tells us that the elevator is moving with a downward acceleration of $0.5\ m/s^2$.

4.

Explanation:

Step1: Draw free - body diagrams

For the upward part of the journey: The forces acting on the person are the gravitational force $F_g = mg$ (downward) and the normal force $N$ (upward). For the downward part of the journey: The forces are still the gravitational force $F_g=mg$ (downward) and the normal force $N$ (upward).

Step2: Compare

During the upward journey, if the elevator is accelerating upward, $N>mg$; if it is moving at a constant speed, $N = mg$; if it is decelerating upward, $N<mg$. During the downward journey, if the elevator is accelerating downward, $N<mg$; if it is moving at a constant speed, $N = mg$; if it is decelerating downward, $N>mg$.

5.

Explanation:

Step1: Apply Newton's second - law

$F_{net}=ma$, and $F_{net}=F_{applied}-0$ (since on ice - skating rink, we assume no friction). Given $F_{applied}=90\ N$ and $m = 65\ kg$.

Step2: Calculate acceleration

$a=\frac{F_{net}}{m}=\frac{F_{applied}}{m}$. Substituting the values, $a=\frac{90}{65}\approx1.38\ m/s^2$.

6.

Explanation:

Step1: Consider Newton's second law for each crate

Let the force applied to each crate be $F$. The mass of each crate is $m=\frac{W}{g}$, where $W$ is the weight of the crate. If we assume we want to move the crate with a constant speed (so $F_{net}=0$ on a low - resistance roller belt), the force applied should balance any small resistive forces. According to Newton's third law, when we push the crate, the crate exerts an equal and opposite force on us. To move the crate onto the truck, we can use an inclined plane. By using an inclined plane, the force required to lift the crate vertically is spread over a longer distance along the incline. According to Newton's second law $F = ma$, if we want to accelerate the crate, we need to apply a force greater than the resistive forces. Also, when we weigh the crate before moving it, the scale reading gives the weight of the crate. When we push the crate, we are applying a force to change its state of motion, either to start it moving or to accelerate it.

Answer: 2. On Earth, the force holding the 10 - kg rock is 98 N. On the Moon, it is approximately 16.3 N. 3. The elevator is accelerating downward at $0.5\ m/s^2$. 4. Free - body diagrams show different normal and gravitational force relationships depending on the motion (acceleration, constant speed, deceleration) of the elevator during upward and downward journeys. 5. The resulting acceleration is approximately $1.38\ m/s^2$. 6. We can use an inclined plane to move the crate onto the truck. When pushing, we apply a force to change its state of motion, and Newton's laws govern the interaction between the crate and us, as well as the measurement of its weight before moving.