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Question 4 of 949

A ball of mass 0.1kg is thrown vertically upwards with a speed of 10ms-1 from the top of a tower 10m high. Neglecting air resistance, its total energy just before hitting the ground is

(take g = 10ms-2)

  • A. 5 J
  • B. 10 J
  • C. 15 J
  • D. 20 J

Correct Answer: C

Explanation
To determine the total energy of the ball just before it hits the ground, we need to consider both its potential energy and kinetic energy at that point. Let's break this down step-by-step. ### Step 1: Understanding Energy Types 1. **Potential Energy (PE)**: This is the energy stored due to an object's position. For an object at height \( h \), the potential energy is given by the formula: \[ PE = mgh \] where: - \( m \) = mass of the object (in kg) - \( g \) = acceleration due to gravity (in m/s²) - \( h \) = height above the reference point (in meters) 2. **Kinetic Energy (KE)**: This is the energy of motion. For an object with mass \( m \) moving at speed \( v \), the kinetic energy is given by the formula: \[ KE = \frac{1}{2} mv^2 \] ### Step 2: Calculate Initial Energies The ball is thrown upwards from a height of 10 m with an initial speed of 10 m/s. We will calculate the potential energy at the top of the tower and the kinetic energy at that point. 1. **Potential Energy at the top of the tower**: \[ PE = mgh = 0.1 \, \text{kg} \times 10 \, \text{m/s}^2 \times 10 \, \text{m} = 10 \, \text{J} \] 2. **Kinetic Energy at the top of the tower**: \[ KE = \frac{1}{2} mv^2 = \frac{1}{2} \times 0.1 \, \text{kg} \times (10 \, \text{m/s})^2 = \frac{1}{2} \times 0.1 \times 100 = 5 \, \text{J} \] ### Step 3: Total Energy at the Top The total mechanical energy (E) at the top of the tower is the sum of the potential and kinetic energy: \[ E = PE + KE = 10 \, \text{J} + 5 \, \text{J} = 15 \, \text{J} \] ### Step 4: Energy Conservation Since we are neglecting air resistance, the total mechanical energy of the ball remains constant throughout its motion. Therefore, just before the ball hits the ground, its total energy will still be 15 J. ### Step 5: Conclusion Thus, the total energy of the ball just before it hits the ground is: \[ \text{Total Energy} = 15 \, \text{J} \] ### Explanation of Options - **Option A (5 J)**: This is incorrect because it only considers the kinetic energy at the top and ignores the potential energy. - **Option B (10 J)**: This is incorrect as it only considers the potential energy at the top and ignores the kinetic energy. - **Option C (15 J)**: This is correct as it accounts for both the potential and kinetic energy at the top of the tower. - **Option D (20 J)**: This is incorrect as it overestimates the total energy by not correctly summing the energies. ### Revision Summary - Total mechanical energy is conserved in the absence of air resistance. - Calculate potential energy using \( PE = mgh \) and kinetic energy using \( KE = \frac{1}{2} mv^2 \). - The total energy at the top of the tower is the sum of potential and kinetic energy. - The total energy just before hitting the ground remains the same as at the top, which is 15 J in this case.
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