Numerical Physics Problems: Force, Mass, Acceleration, Pressure, and Gravitational Calculations

Mass Determination from Weight:
- The relationship between weight ( $W$ ), mass ( $m$ ), and acceleration due to gravity ( $g$ ) is defined by the formula: $W = m \times g$
- Rearranging the equation to solve for mass ( $m$ ) gives: $m = \frac{W}{g}$
- Application to given values:
  - Weight: $W = 50\,N$
  - Acceleration due to gravity: $g = 10\,m\,s^{-2}$
  - Mass representation: $x\,kg$
  - Calculation: $x = \frac{50\,N}{10\,m\,s^{-2}} = 5$
  - The calculated mass $x$ is $5$ , corresponding to $5\,kg$ .
Weight Determination from Mass:
- The weight of a body is calculated by multiplying its mass by the gravitational acceleration: $W = m \times g$
- Application to given values:
  - Mass of the body: $m = 60\,kg$
  - Acceleration due to gravity: $g = 10\,m\,s^{-2}$
  - Target form for weight: $x \times 10^2\,N$
  - Calculation: $W = 60\,kg \times 10\,m\,s^{-2} = 600\,N$
  - Converting $600\,N$ into scientific format: $600\,N = 6 \times 10^2\,N$
  - Equating $6 \times 10^2\,N$ to $x \times 10^2\,N$ gives $x = 6$

Pressure Calculation on a Surface of a Cube:
- Pressure ( $P$ ) is defined as force ( $F$ ) per unit area ( $A$ ): $P = \frac{F}{A}$
- For a cube with edge length $a$ , the surface area $A$ of one face is given by: $A = a^2$
- Application to given values:
  - Uniformly distributed force: $F = 16\,N$
  - Edge length of the cube: $a = 4\,cm = 4 \times 10^{-2}\,m$
  - Target pressure representation: $x \times 10^4\,Pa$
  - Area calculation: $A = (4 \times 10^{-2}\,m)^2 = 16 \times 10^{-4}\,m^2$
  - Substitution into the pressure formula: $P = \frac{16\,N}{16 \times 10^{-4}\,m^2} = 10^4\,Pa$
  - Expressing $10^4\,Pa$ in terms of $x \times 10^4\,Pa$ : $1 \times 10^4\,Pa = x \times 10^4\,Pa$
  - The resulting value of $x$ is $1$

Acceleration of a Mass on a Frictionless Surface:
- According to Newton's Second Law of Motion, force is the product of mass and acceleration: $F = m \times a$
- Solving for acceleration ( $a$ ): $a = \frac{F}{m}$
- Application to given values:
  - Applied horizontal force: $F = 4\,N$
  - Mass of the block: $m = 2\,kg$
  - Surface condition: Frictionless (net force equals applied force)
  - Calculation: $a = \frac{4\,N}{2\,kg} = 2\,m\,s^{-2}$
  - The acceleration of the block is $2\,m\,s^{-2}$ .
Net Force from Opposing Odd Consecutive Forces:
- Odd consecutive forces can be represented generalistically as $F_1 = 2n + 1$ and $F_2 = 2n + 3$ , where $n$ is an integer.
- When two horizontal forces $F_1$ and $F_2$ act on a body in opposite directions, the net force ( $F_{net}$ ) is the absolute magnitude of their difference: $F_{net} = |F_2 - F_1$
- Calculation for consecutive odd forces: $F_{net} = |(2n + 3) - (2n + 1)| = 2\,N$
- The net force acting on the body is $2\,N$ .