Katex test post
Published: 14 Mar 2026
Which formula represents work (W) when force (F) moves an object over a displacement (S)? [katex]W = F \times S[/katex] [katex]W = \frac{F}{S}[/katex] [katex]W = F + S[/katex] [katex]W = F – S[/katex] [katex]W = F \times S[/katex] Work is calculated as the product of force and displacement in the direction of the force. How is work calculated when a force is applied at an angle to the direction of motion? [katex]W = F \times S[/katex] [katex]W = \frac{F}{S}[/katex] [katex]W = F \times S \times \cos(\theta)[/katex] [katex]W = F \times S \times \sin(\theta)[/katex] [katex]W = F \times S \times \cos(\theta)[/katex] The work done when a force is applied at an angle is calculated using the formula [katex]W = F \times S \times \cos(\theta)[/katex]. What is the mathematical expression for gravitational potential energy? [katex]P.E = m \times g \times h[/katex] [katex]P.E = F \times s[/katex] [katex]P.E = mv^{2}[/katex] [katex]P.E = E \times t[/katex] [katex]P.E = m \times g \times h[/katex] Gravitational potential energy is calculated using the formula [katex]P.E = m \times g \times h[/katex]. What is the work done against the force of gravity when lifting an object? [katex]W = F \times d[/katex] [katex]W = m \times g \times h[/katex] [katex]W = P \times t[/katex] [katex]W = KE + PE[/katex] [katex]W = m \times g \times h[/katex] When lifting an object against gravity, the work done is given by the formula [katex]W = m \times g \times h[/katex].
Which formula represents work (W) when force (F) moves an object over a displacement (S)?
An antilogarithm (commonly shortened to antilog) is the exact reverse or inverse operation of finding a logarithm.
If taking a logarithm answers the question: “To what power must I raise the base to get this number?”, then taking the antilogarithm answers the reverse question: “What was the original number?”
Core Concept: Logarithm vs. Antilogarithm
When working with base $10$ (common logarithms):
$$\text{If } \log_{10}(x) = y, \quad \text{then } \text{antilog}(y) = x = 10^y$$
- Logarithm: Takes a number and tells you the power.
- Since $10^2 = 100$, we write $\log_{10}(100) = 2$.
- Antilogarithm: Takes the power and gives you back the original number.
- $\text{antilog}(2) = 10^2 = 100$.
The Two Parts of a Logarithm
Every logarithm value has two components:
- Characteristic: The whole integer part (to the left of the decimal). It can be positive, zero, or negative. It dictates where to place the decimal point.
- Mantissa: The fractional decimal part (to the right of the decimal). For standard table use, the mantissa must always be positive. It dictates the sequence of significant digits.
Example: In the number $2.4955$:
- Characteristic = $2$
- Mantissa = $.4955$
Step-by-Step Method to Find Antilog (Using Log/Antilog Tables)
To evaluate an antilog like $\text{antilog}(2.4955)$:
- Separate the parts:
- Characteristic = $2$
- Mantissa = $.4955$
- Look up the Mantissa in the Antilog Table:
- Look down the leftmost column to find $.49$.
- Move horizontally across to the column under $5$$\rightarrow$ Read the 4-digit number (e.g., $3126$).
- In the same row, move to the Mean Difference section under column $5$$\rightarrow$ Read the difference value (e.g., $4$).
- Add the mean difference to your number: $3126 + 4 = 3130$.
- Use the Characteristic to Place the Decimal Point:
- Rule: If the characteristic is $c$ (where $c \ge 0$), place the decimal point so there are $(c + 1)$ digits before the decimal.
- Here, $c = 2$, so there must be $2 + 1 = 3$ digits before the decimal:
$$\text{antilog}(2.4955) \approx 313.0$$
Handling Negative Characteristics (Bar Notation)
In school math, negative characteristics are written with a bar over the number to show that only the characteristic is negative, while the mantissa remains positive.
Example: $\text{antilog}(\bar{2}.4955)$
- Characteristic = $-2$ (written as $\bar{2}$)
- Mantissa = $.4955$
- Find the digits using the mantissa:
- Using $.4955$ in the table gives the exact same digits: $3130$.
- Apply the decimal rule for negative characteristic ($\bar{n}$):
- Place $(n – 1)$ zeros immediately after the decimal point before writing the digits.
- Here, $n = 2$, so we insert $2 – 1 = 1$ zero:$$\text{antilog}(\bar{2}.4955) \approx 0.03130$$
Summary Rules for Decimal Placement
| Characteristic | Meaning | Decimal Rule | Example Digits (3130) | Result |
| $0$ | $10^0$ | $0 + 1 = 1$ digit before decimal | $3.130$ | $3.130$ |
| $1$ | $10^1$ | $1 + 1 = 2$ digits before decimal | $31.30$ | $31.30$ |
| $2$ | $10^2$ | $2 + 1 = 3$ digits before decimal | $313.0$ | $313.0$ |
| $\bar{1}$ ($-1$) | $10^{-1}$ | No leading zeros after decimal | $0.3130$ | $0.3130$ |
| $\bar{2}$ ($-2$) | $10^{-2}$ | $1$ zero after decimal | $0.03130$ | $0.03130$ |
| $\bar{3}$ ($-3$) | $10^{-3}$ | $2$ zeros after decimal | $0.003130$ | $0.003130$ |
[katex]W = F \times S[/katex]
[katex]W = \frac{F}{S}[/katex]
[katex]W = F + S[/katex]
[katex]W = F – S[/katex]
[katex]W = F \times S[/katex]
Work is calculated as the product of force and displacement in the direction of the force. How is work calculated when a force is applied at an angle to the direction of motion?
[katex]W = F \times
S[/katex]
[katex]W = \frac{F}{S}[/katex]
[katex]W = F \times S
\times \cos(\theta)[/katex]
[katex]W = F \times S
\times \sin(\theta)[/katex]
[katex]W = F \times S
\times \cos(\theta)[/katex]
The work done when a force is applied at an angle is calculated using the formula [katex]W = F \times S \times \cos(\theta)[/katex]. What is the mathematical expression for gravitational potential energy?
[katex]P.E = m \times g
\times h[/katex]
[katex]P.E = F \times
s[/katex]
[katex]P.E = mv^{2}[/katex]
[katex]P.E = E \times
t[/katex]
[katex]P.E = m \times g
\times h[/katex]
Gravitational potential energy is calculated using the formula [katex]P.E = m \times g \times h[/katex]. What is the work done against the force of gravity when lifting an object?
[katex]W = F \times
d[/katex]
[katex]W = m \times g
\times h[/katex]
[katex]W = P \times
t[/katex]
[katex]W = KE + PE[/katex]
[katex]W = m \times g
\times h[/katex]
When lifting an object against gravity, the work done is given by the formula [katex]W = m \times g \times h[/katex].
This is Symbol [katex] a^2+bx \% =-5 [/katex]
Percentage symbol % shows
This is [katex] \dfrac{2}{5} \times \frac{2}{5} % [/katex]This is [katex] \dfrac{2}{5} \times \frac{2}{5} % [/katex]
This is [katex] \frac{2}{5} [/katex]
[katex] \begin{array}{l|rr} 2 & 8, & 12 \\ \hline 2 & 4, & 6 \\ \hline 2 & 2, & 3 \\ \hline 3 & 1, & 3 \\ \hline & 1, & 1 \end{array} [/katex]This is [katex] \dfrac{2}{5} \dfrac{2}{5} [/katex]
This is [katex] \dfrac{2}{5} [/katex]
This is [katex] \frac{2}{5} [/katex]
This is [katex] \frac{2}{5} \cdot \frac{2}{5} [/katex]
This is [katex] \frac{2}{5} [/katex]

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- Be Respectful
- Stay Relevant
- Stay Positive
- True Feedback
- Encourage Discussion
- Avoid Spamming
- No Fake News
- Don't Copy-Paste
- No Personal Attacks

