Problem 78 - University Physics with Modern Physics

Question

One end of a horizontal spring with force constant 76.0 N/m is attached to a vertical post. A 2.00-kg block of frictionless ice is attached to the other end and rests on the floor. The spring is initially neither stretched nor compressed. A constant horizontal force of 54.0 N is then applied to the block, in the direction away from the post. (a) What is the speed of the block when the spring is stretched 0.400 m? (b) At that instant, what are the magnitude and direction of the acceleration of the block?

Step-by-Step Solution

Answer

(a) Speed is 4.39 m/s. (b) Acceleration is 11.8 m/s² away from the post.

Step 1: Understand the Problem

We have a spring-block system where the spring constant is 76.0 N/m, and the block has a mass of 2.00 kg. A force of 54.0 N is applied to the block. We need to find the speed of the block when the spring is stretched by 0.400 m, and also determine the acceleration at that instant.

Step 2: Apply Work-Energy Principle

The work done on the block by the external force goes into kinetic energy and the potential energy stored in the spring. Use the equation: \( W_{ext} = \Delta KE + \Delta PE \), where \( W_{ext} = F \times d \), \( \Delta KE = \frac{1}{2} m v^2 \), and \( \Delta PE = \frac{1}{2} k x^2 \).

Step 3: Calculate Work Done by External Force

The work done by the external force is \( W = F \cdot d = 54.0 \times 0.400 \). Calculating this gives \( W = 21.6 \) J.

Step 4: Calculate Potential Energy in the Spring

The potential energy stored in the spring when it is stretched by 0.400 m is \( PE = \frac{1}{2} k x^2 = \frac{1}{2} \times 76.0 \times (0.400)^2 \). This results in \( PE = 6.08 \) J.

Step 5: Solve for the Velocity of the Block

Using the work-energy principle, we have \( 21.6 = \frac{1}{2} \times 2.00 \times v^2 + 6.08 \). Simplifying gives \( 21.6 - 6.08 = v^2 \). Solve for \( v \) to find \( v = \sqrt{(21.6 - 6.08)/1} = 4.39 \) m/s.

Step 6: Calculate the Acceleration of the Block

Use Newton's second law: \( F_{net} = m a \). Here, \( F_{net} = F - kx \) because both the external force and the spring force act in opposite directions. Thus, \( F_{net} = 54.0 - 76.0 \times 0.400 \). This results in \( F_{net} = 23.6 \) N. Then, \( a = \frac{F_{net}}{m} = \frac{23.6}{2.00} = 11.8 \) m/s².

Key Concepts

Work-Energy Principle

The work-energy principle is like a balance sheet for energy. It states that the work done on an object is equal to the change in its kinetic and potential energy.
In the context of our spring-block system, when a constant force is applied, it stretches the spring and gives energy to the block.
Energy is not lost; it's just converted from one form to another. Here's how it works in our exercise:

• The work done by the external force is calculated by multiplying the force by the distance it moves the block: \( W_{ext} = F \times d \).
• Kinetic energy \( (KE) \) is the energy of motion, given by \( \Delta KE = \frac{1}{2} m v^2 \). This is the energy the block gains as it starts to move faster.
• Potential energy \( (PE) \) is stored energy due to the spring's stretch, given by \( \Delta PE = \frac{1}{2} k x^2 \).

By balancing these energies, we use the equation: \( W_{ext} = \Delta KE + \Delta PE \), to solve for the block's speed. It's like following the trail of energy changes to understand movement in the system.

Potential Energy

Potential energy in a spring-block system can be thought of as the spring's capacity to do work when released. When the spring is stretched or compressed, it stores energy, akin to winding a rubber band.
In our problem:

• The spring's potential energy depends on how far it is stretched or compressed. We use the formula \( PE = \frac{1}{2} k x^2 \), where \( k \) is the spring constant (76.0 N/m), and \( x \) is the stretch (0.400 m in this case).
• The result of this calculation for our exercise is \( 6.08 \) J (joules), representing energy stored in the spring.
• This stored energy can be converted back into kinetic energy when the spring returns to its original position.

Understanding potential energy helps us predict and calculate how the spring will affect the block's motion, thus playing a crucial role in analyzing motion.

Newton's Second Law

Newton's second law is all about relating forces, mass, and acceleration, forming the bedrock of dynamics. It states that the net force acting on an object equals its mass times its acceleration: \( F_{net} = m a \). In our spring-block system:

• The block is subjected to two main forces: the external applied force and the opposing spring force.
• The net force \( F_{net} \) is calculated by taking the difference between the applied force and the spring force: \( F_{net} = F - kx \), where \( F = 54.0 \) N and \( kx = 76.0 \times 0.400 \) N.
• This gives us \( F_{net} = 23.6 \) N, which we use to find the block’s acceleration through \( a = \frac{F_{net}}{m} \).
• The result is an acceleration of \( 11.8 \) m/s², indicating how quickly the block speeds up as it moves away from the equilibrium point.

By understanding Newton's second law, we can determine not just how fast an object moves, but also how it accelerates or decelerates as forces in the system change.