Question

Ex 4: A block is attached to one end of a horizontal spring with a constant k = 200 N/m. The block is pulled a distance of 0.40 m away from equilibrium and released so that it slides back and forth along a frictionless surface. When the block is 0.20 m away from equilibrium, it is traveling with a speed of 0.5 m/s. What is the mass of the block?

Answer 1

Answer 2

Given data:

• Spring constant: $k = 200$ N/m

• Maximum displacement (amplitude): $A = 0.40$ m

• Displacement at given speed: $x = 0.20$ m

• Speed at $x = 0.20$ m: $v = 0.5$ m/s

Step 1: Total Energy in the System

Total mechanical energy in a spring system is given by:

$$E = \frac{1}{2} k A^2$$ 
$$E = \frac{1}{2} (200) (0.40)^2$$ 
$$E = 16 \text{ J}$$

Step 2: Energy at $x = 0.20$ m

At $x = 0.20$ m, the energy is split between potential energy $U$ and kinetic energy $K$.

$$U = \frac{1}{2} k x^2$$ 
$$U = \frac{1}{2} (200) (0.20)^2$$ 
$$U = 4 \text{ J}$$ 

Step 3: Find the Kinetic Energy

Since total energy is conserved:

$$K = E - U$$
$$K = 16 - 4 = 12 \text{ J}$$

Kinetic energy is also given by:

$$K = \frac{1}{2} m v^2$$ 
$$12 = \frac{1}{2} m (0.5)^2$$ 
$$12 = \frac{1}{2} m (0.25)$$ 
$$m = \frac{12 \times 2}{0.25} = 96 \text{ kg}$$ 

 Answer is :-

$$\boxed{{\displaystyle \mathrm{}}}$$$$\boxed{{\displaystyle 96\text{ kg}}}$$$$\boxed{96 \text{ kg}}$$

Answer 3

Let's solve this problem using the conservation of energy principle.

1. Understand the Energy in the System

• Initial Potential Energy (PE_initial): When the block is pulled 0.40 m from equilibrium, all energy is stored as potential energy in the spring.

• Energy at 0.20 m (PE_final + KE_final): When the block is 0.20 m from equilibrium, some energy is still stored as potential energy in the spring, and the rest is kinetic energy of the moving block.

2. Write the Energy Equations

• Initial Potential Energy (PE_initial):

  PE_initial = (1/2) * k * x_initial²

  where k = 200 N/m and x_initial = 0.40 m

• Final Potential Energy (PE_final):

  PE_final = (1/2) * k * x_final²

  where x_final = 0.20 m

• Final Kinetic Energy (KE_final):

  KE_final = (1/2) * m * v²

  where m is the mass we want to find and v = 0.5 m/s

3. Apply Conservation of Energy

Since there's no friction, the total energy is conserved:

PE_initial = PE_final + KE_final

4. Plug in the Values and Solve for Mass (m)

(1/2) * 200 N/m * (0.40 m)² = (1/2) * 200 N/m * (0.20 m)² + (1/2) * m * (0.5 m/s)²

100 * 0.16 = 100 * 0.04 + (1/2) * m * 0.25

16 = 4 + 0.125 * m

12 = 0.125 * m

m = 12 / 0.125

m = 96 kg

Answer:

The mass of the block is 96 kg.