JEE-Main

1. Normal force = Weight of block - Force applied perpendicular to the surface

$$N=mg-F\sin \theta$$

$$N=(1\text{ kg})(9.81{\text{ m/s}}^2)-(10\text{ N})(\sin 30\text{degree})$$

$$N=\boxed{{\displaystyle 8.54\text{ N}}}$$

2. Maximum height reached by a projectile:

$$h=\frac{v_i^2{\sin }^2\theta }{2g}$$

Where:

• (h) is the maximum height reached (in meters)
• (v_i) is the initial velocity (in m/s)
• (\theta) is the angle of projection (in degrees)
• (g) is the acceleration due to gravity (9.81 m/s^2)

Substituting the given values:

$$h=\frac{(10\text{ m/s}{)}^2{\sin }^{2}90\text{degree}}{2(9.81{\text{ m/s}}^2)}$$

$$h=\boxed{{\displaystyle 4.58\text{ m}}}$$

3. Work done by a spring:

$$W=\frac{1}{2}k{x}^2$$

Where:

• (W) is the work done (in Joules)
• (k) is the spring constant (in N/m)
• (x) is the displacement of the spring (in meters)

Substituting the given values:

$$W=\frac{1}{2}(100\text{ N/m})(0.1\text{ m}{)}^2$$

$$W=\boxed{{\displaystyle 0.5\text{ J}}}$$

4. Gravitational force acting on a satellite:

$$F_g=\frac{G{m}_1{m}_2}{r^2}$$

Where:

• (F_g) is the gravitational force (in Newtons)
• (G) is the gravitational constant (6.674 × 10-11 N·m^2/kg^2)
• (m_1) is the mass of the earth (5.972 × 10^24 kg)
• (m_2) is the mass of the satellite (100 kg)
• (r) is the distance between the center of the earth and the satellite (6.371 × 10^6 m + 400 × 10^3 m)

Substituting the given values:

$$F_g=\frac{(6.674\times {10}^{-11}{\text{ N·m}}^2/{\text{kg}}^2)(5.972\times {10}^{24}\text{ kg})(100\text{ kg})}{(6.371\times {10}^6\text{ m}+400\times {10}^3\text{ m}{)}^2}$$

$$F_g=\boxed{{\displaystyle 5.93\times {10}^2\text{ N}}}$$

CBSE-Boards

1. Normal force acting on a book:

$$N=mg$$

Where:

• (N) is the normal force (in Newtons)
• (m) is the mass of the book (in kilograms)
• (g) is the acceleration due to gravity (9.81 m/s^2)

Substituting the given values:

$$N=(1\text{ kg})(9.81{\text{ m/s}}^2)$$

$$N=\boxed{{\displaystyle 9.81\text{ N}}}$$

2. Maximum height reached by a projectile:

$$h=\frac{v_i^2{\sin }^2\theta }{2g}$$

Where:

• (h) is the maximum height reached (in meters)
• (v_i) is the initial velocity (in m/s)
• (\theta) is the angle of projection (in degrees)
• (g) is the acceleration due to gravity (9.81 m/s^2)

Substituting the given values:

$$h=\frac{(5\text{ m/s}{)}^2{\sin }^{2}90\text{degree}}{2(9.81{\text{ m/s}}^2)}$$

$$h=\boxed{{\displaystyle 1.27\text{ m}}}$$

3. Work done by a spring:

$$W=\frac{1}{2}k{x}^2$$

Where:

• (W) is the work done (in Joules)
• (k) is the spring constant (in N/m)
• (x) is the displacement of the spring (in meters)

Substituting the given values:

$$W=\frac{1}{2}(50\text{ N/m})(0.05\text{ m}{)}^2$$

$$W=\boxed{{\displaystyle 0.0625\text{ J}}}$$

4. Gravitational force acting on a satellite:

$$F_g=\frac{G{m}_1{m}_2}{r^2}$$

Where:

• (F_g) is the gravitational force (in Newtons)
• (G) is the gravitational constant (6.674 × 10-11 N·m^2/kg^2)
• (m_1) is the mass of the earth (5.972 × 10^24 kg)
• (m_2) is the mass of the satellite (50 kg)
• (r) is the distance between the center of the earth and the satellite (6.371 × 10^6 m + 200 × 10^3 m)

Substituting the given values:

$$F_g=\frac{(6.674\times {10}^{-11}{\text{ N·m}}^2/{\text{kg}}^2)(5.972\times {10}^{24}\text{ kg})(50\text{ kg})}{(6.371\times {10}^6\text{ m}+200\times {10}^3\text{ m}{)}^2}$$

$$F_g=\boxed{{\displaystyle 2.96\times {10}^2\text{ N}}}$$