Question

A rocket of initial mass 125 kg (including all the contents) has an engine that produces a constant vertical force (the \(thrust\)) of 1720 N. Inside this rocket, a 15.5-N electrical power supply rests on the floor. (a) Find the initial acceleration of the rocket. (b) When the rocket initially accelerates, how hard does the floor push on the power supply? (\(Hint:\) Start with a free- body diagram for the power supply.)

Short answer

(a) The initial acceleration of the rocket is 3.96 m/s². (b) The floor pushes on the power supply with a force of 21.8 N.

Step-by-step solution

Understanding Forces Acting on the Rocket

The thrust produced by the engine is 1720 N, and this acts upwards against gravity and provides the acceleration. The weight of the rocket acts downwards. We use the force equation: \[ F_{ ext{net}} = F_{ ext{thrust}} - F_{ ext{weight}} \] where \( F_{\text{weight}} = m_{\text{rocket}} \times g \), with \( g = 9.8 \, \text{m/s}^2 \).

Calculate Weight of the Rocket

Calculate the gravitational force acting on the rocket using: \[ F_{\text{weight}} = m_{\text{rocket}} \times g = 125 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 1225 \, \text{N} \]

Determine Net Force and Acceleration of the Rocket

Using the net force equation from Step 1, calculate the net force:\[ F_{\text{net}} = F_{\text{thrust}} - F_{\text{weight}} = 1720 \, \text{N} - 1225 \, \text{N} = 495 \, \text{N} \]Then, use Newton's second law to find the acceleration:\[ F_{\text{net}} = m_{\text{rocket}} \times a \Rightarrow a = \frac{F_{\text{net}}}{m_{\text{rocket}}} = \frac{495 \, \text{N}}{125 \, \text{kg}} = 3.96 \, \text{m/s}^2 \]

Free-Body Diagram of Power Supply

The power supply experiences two forces: its weight \( W = 15.5 \, \text{N} \) acting downwards and the normal force exerted by the floor acting upwards. Since the power supply is accelerating with the rocket, apply Newton's second law to it.

Calculate Normal Force on the Power Supply

Use the following formula for the forces on the power supply:\[ F_{\text{normal}} - W = m_{\text{power}} \times a \] Where \( m_{\text{power}} \) is the mass of the power supply:\[ m_{\text{power}} = \frac{W}{g} = \frac{15.5 \, \text{N}}{9.8 \, \text{m/s}^2} = 1.58 \, \text{kg} \]Now find the normal force:\[ F_{\text{normal}} = W + m_{\text{power}} \times a = 15.5 \, \text{N} + 1.58 \, \text{kg} \times 3.96 \, \text{m/s}^2 = 21.8 \, \text{N} \]

Key concepts

Newton's Second Law

Newton's Second Law is a vital part of physics, especially when calculating the motion of objects under the influence of forces. It states that the acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass. The mathematical representation is:\[ F_{\text{net}} = m \times a \]where \( F_{\text{net}} \) is the net force, \( m \) is the mass of the object, and \( a \) is the acceleration.Understanding this law helps us explain how forces result in changes to the motion of an object. When applied to the rocket in our exercise, this law allows us to determine how the thrust affects its acceleration. After finding the net force from thrust and weight, dividing that by the rocket's mass gives us the acceleration. This calculation showcases how increasing the force or decreasing the mass impacts the rocket's acceleration, making it a fundamental concept in rocket science.

Free-Body Diagram

A free-body diagram is a simple drawing that shows all the forces acting on an object. This visual tool is crucial for understanding and analyzing the dynamics of objects. To draw a free-body diagram, follow these steps:

• Identify the object you're focusing on.
• Draw the object as a simple shape, like a box or a dot.
• Draw arrows representing all the forces acting on the object; the direction of each arrow shows the force's direction, and the arrow's length indicates the force's magnitude.

In our exercise, we created a free-body diagram for both the rocket and the power supply. For the rocket, the diagram included the force of thrust acting upwards and the weight acting downwards. For the power supply, it involved its weight and the normal force from the floor. Such diagrams simplify complex problems by breaking them into manageable parts, which makes it easier to apply physical laws, like Newton's Second Law, for solutions.

Gravitational Force

Gravitational force is the attractive force that acts between any two masses. On Earth's surface, it gives weight to physical objects and governs the motion of celestial bodies. The gravitational force on an object is calculated by:\[ F_{\text{gravity}} = m \times g \]where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \) on Earth).In our rocket exercise, gravitational force played a crucial role, both in calculating the rocket's weight and the weight of the power supply. These weights represent the force exerted by gravity, which opposes the rocket's thrust and affects the net force and acceleration calculations. Understanding gravity's role is essential, as it is a key factor in balancing the forces acting on objects, especially in cases involving vertical movement, such as in rocketry.

Normal Force

Normal force is the support force exerted upon an object that is in contact with another stable surface. When an object is resting on a horizontal surface, it is the force exerted by that surface to support the weight of the object.In our problem, the normal force acts on the power supply from the floor of the rocket. The calculation involves

• Identifying the weight of the power supply.
• Adding the additional force due to the acceleration of the rocket (since both the rocket and power supply accelerate together).

Therefore, we calculated the normal force using:\[ F_{\text{normal}} = W + m_{\text{power}} \times a \]where \( W \) is the weight of the power supply, and \( a \) is the common acceleration. Understanding how normal force balances other forces like gravity ensures equilibrium within a system, despite movement or acceleration. This is particularly important in maintaining integrity in structural scenarios, such as when items inside a rocket need to remain stable during acceleration.