if a 10 kg weight has 10 newtons applied at 45 degrees, how to get its trajectory?

To determine the trajectory of a 10 kg weight subjected to a 10 N force applied at a 45-degree angle, we'll need to calculate the horizontal and vertical components of the force, then use Newton's second law of motion to determine the weight's acceleration and trajectory. Here's a step-by-step approach:

### 1. Break the force into components:
The force of 10 N is applied at a 45-degree angle, so we can resolve this force into two components: horizontal (\( F_x \)) and vertical (\( F_y \)).

\[
F_x = F \cdot \cos(45^\circ) = 10 \cdot \frac{1}{\sqrt{2}} = \frac{10}{\sqrt{2}} \approx 7.07 \, \text{N}
\]
\[
F_y = F \cdot \sin(45^\circ) = 10 \cdot \frac{1}{\sqrt{2}} = \frac{10}{\sqrt{2}} \approx 7.07 \, \text{N}
\]

### 2. Calculate the acceleration:
Now, using Newton’s second law (\( F = m \cdot a \)), we can find the acceleration in both directions. The mass of the object is 10 kg.

#### Horizontal acceleration:
\[
a_x = \frac{F_x}{m} = \frac{7.07}{10} \approx 0.707 \, \text{m/s}^2
\]

#### Vertical acceleration:
We need to account for the force of gravity in the vertical direction. The gravitational force acting on the object is:

\[
F_{\text{gravity}} = m \cdot g = 10 \cdot 9.8 = 98 \, \text{N}
\]

Thus, the net vertical force is:

\[
F_{\text{net, y}} = F_y - F_{\text{gravity}} = 7.07 - 98 = -90.93 \, \text{N}
\]

This gives the vertical acceleration:

\[
a_y = \frac{F_{\text{net, y}}}{m} = \frac{-90.93}{10} \approx -9.093 \, \text{m/s}^2
\]

### 3. Determine the trajectory:
To determine the motion of the weight, we'll use kinematic equations. For simplicity, assume the weight starts from rest and the initial position is at the origin.

#### Horizontal motion:
The horizontal displacement over time \( t \) is given by:

\[
x(t) = \frac{1}{2} a_x t^2 = \frac{1}{2} \cdot 0.707 \cdot t^2 = 0.3535 t^2 \, \text{m}
\]

#### Vertical motion:
The vertical displacement over time \( t \) is:

\[
y(t) = \frac{1}{2} a_y t^2 = \frac{1}{2} \cdot (-9.093) \cdot t^2 = -4.5465 t^2 \, \text{m}
\]

### 4. Equation of the trajectory:
Now, eliminate \( t \) to find the trajectory equation. From \( x(t) = 0.3535 t^2 \), solve for \( t^2 \):

\[
t^2 = \frac{x}{0.3535}
\]

Substitute this into the equation for \( y(t) \):

\[
y = -4.5465 \cdot \frac{x}{0.3535} = -12.86x
\]

Thus, the equation of the trajectory is approximately:

\[
y = -12.86x
\]

### Summary:
The trajectory of the weight will be a straight line downward with a slope of -12.86. The horizontal motion is uniformly accelerated, and the vertical motion is dominated by gravity.

Let me know if you'd like more details or have any questions!

Here are five related questions for further exploration:
1. How does changing the angle of the applied force affect the trajectory?
2. What happens if the object is not starting from rest?
3. How would the trajectory change if air resistance is considered?
4. What would the trajectory look like if the force is applied in pulses instead of continuously?
5. How can you determine the time of flight for a projectile motion?

**Tip**: For objects under force at an angle, breaking the force into components simplifies the analysis.

Learn how to calculate the trajectory of a 10 kg weight subjected to a 10 N force at a 45-degree angle. This problem explores vector decomposition, Newton's laws, and kinematic equations to determine the object's motion. Suitable for high school students in Grades 9-12.

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