Class 9 - Unit # 8 : Energy Sources and Transfer of Energy - Solved Numericals

Work

2. How much work is needed to move horizontally a body $20\text{ m}$ by a force of $30\text{ N}$, the angle between the body and the horizontal surface is $60^\circ$?

Data:

• Applied Force ($F$) = $30\text{ N}$
• Displacement ($d$) = $20\text{ m}$
• Angle with Horizontal ($\theta$) = $60^\circ$
• Work Done ($W$) = ?

Solution:

According to the definition of work, when a force acts at an angle $\theta$ to the direction of motion:

$$W = F \cdot d \cdot \cos(\theta)$$ $$W = 30\text{ N} \times 20\text{ m} \times \cos(60^\circ)$$ $$W = 600 \times 0.5$$ $$W = 300\text{ J}$$

RESULT: The work needed in this case is $300\text{ J}$.

3. How much work is done, if a crate is moved at a distance of $50\text{ m}$, when a force of $30\text{ N}$ is applied along the surface.

Data:

• Applied Force ($F$) = $30\text{ N}$
• Displacement ($d$) = $50\text{ m}$
• Work Done ($W$) = ?

Solution:

According to the definition of work, since the force is applied along the surface ($\theta = 0^\circ$, so $\cos(0^\circ) = 1$):

$$W = F \cdot d$$ $$W = 30\text{ N} \times 50\text{ m}$$ $$W = 1500\text{ J}$$

RESULT: The work done is $1500\text{ J}$.

Energy Forms

6. What will be the Kinetic energy of a boy of mass $50\text{ kg}$ driving a bike with velocity of $2\text{ m/s}$?

Data:

• Mass ($m$) = $50\text{ kg}$
• Velocity ($v$) = $2\text{ m/s}$
• Kinetic Energy ($\text{K.E.}$) = ?

Solution:

According to the definition of Kinetic Energy:

$$\text{K.E.} = \frac{1}{2} m v^2$$ $$\text{K.E.} = \frac{1}{2} \times 50\text{ kg} \times (2\text{ m/s})^2$$ $$\text{K.E.} = 25 \times 4$$ $$\text{K.E.} = 100\text{ J}$$

RESULT: The kinetic energy of the bike and driver system is $100\text{ J}$.

8. a) If an LED screen of mass $10\text{ kg}$ is lifted up and kept on a cupboard of height $2\text{ m}$. Calculate the potential energy stored in the LED screen.

Data:

• Mass of LED ($m$) = $10\text{ kg}$
• Height lifted ($h$) = $2\text{ m}$
• Acceleration due to gravity ($g$) = $10\text{ m/s}^2$
• Potential Energy ($\text{P.E.}$) = ?

Solution:

According to the definition of Gravitational Potential Energy:

$$\text{P.E.} = m \cdot g \cdot h$$ $$\text{P.E.} = 10\text{ kg} \times 10\text{ m/s}^2 \times 2\text{ m}$$ $$\text{P.E.} = 200\text{ J}$$

RESULT: The potential energy stored in the LED screen is $200\text{ J}$.

8. b) Calculate the potential energy of $3\text{ kg}$ water raised to the tank at the roof of a home $4\text{ m}$ high. (Assume $g = 10\text{ m/s}^2$)

Data:

• Mass of water ($m$) = $3\text{ kg}$
• Height lifted ($h$) = $4\text{ m}$
• Acceleration due to gravity ($g$) = $10\text{ m/s}^2$
• Potential Energy ($\text{P.E.}$) = ?

Solution:

According to the definition of Gravitational Potential Energy:

$$\text{P.E.} = m \cdot g \cdot h$$ $$\text{P.E.} = 3\text{ kg} \times 10\text{ m/s}^2 \times 4\text{ m}$$ $$\text{P.E.} = 120\text{ J}$$

RESULT: The potential energy of the water is $120\text{ J}$.