A 1kg mass moves from point A(0,0) to a point B(1,0).Its speed at point B is 5m/s . If U=2-x^2 and the external agent does 2 J of work on the body, find the initial speed of the body.

To find the initial speed of the body at point A, we can use the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy plus any change in potential energy. In this scenario, we have a mass moving from point A to point B, and we know the work done by an external agent, the potential energy function, and the final speed at point B. Let's break this down step by step.

Understanding the Problem

We have the following information:

• Mass (m) = 1 kg
• Initial position A = (0, 0)
• Final position B = (1, 0)
• Final speed at B (v_B) = 5 m/s
• Work done by the external agent (W) = 2 J
• Potential energy function (U) = 2 - x²

Calculating Potential Energy

First, we need to calculate the potential energy at both points A and B using the given potential energy function:

• At point A (x = 0):

U_A = 2 - (0)² = 2 J

• At point B (x = 1):

U_B = 2 - (1)² = 2 - 1 = 1 J

Change in Potential Energy

The change in potential energy (ΔU) as the mass moves from A to B is:

ΔU = U_B - U_A = 1 J - 2 J = -1 J

Kinetic Energy at Point B

The kinetic energy (KE) at point B can be calculated using the formula:

KE_B = (1/2) * m * v_B²

Substituting the values:

KE_B = (1/2) * 1 kg * (5 m/s)² = (1/2) * 1 * 25 = 12.5 J

Applying the Work-Energy Principle

According to the work-energy principle, the work done on the mass is equal to the change in kinetic energy plus the change in potential energy:

W = ΔKE + ΔU

We can express the change in kinetic energy as:

ΔKE = KE_B - KE_A

Substituting the known values, we get:

2 J = (12.5 J - KE_A) + (-1 J)

Rearranging this gives us:

2 J = 12.5 J - KE_A - 1 J

2 J = 11.5 J - KE_A

KE_A = 11.5 J - 2 J = 9.5 J

Finding the Initial Speed

Now that we have the kinetic energy at point A, we can find the initial speed (v_A) using the kinetic energy formula:

KE_A = (1/2) * m * v_A²

Substituting the values:

9.5 J = (1/2) * 1 kg * v_A²

Multiplying both sides by 2 gives:

19 J = 1 kg * v_A²

Now, dividing by 1 kg:

v_A² = 19 m²/s²

Taking the square root of both sides gives:

v_A = √19 m/s ≈ 4.36 m/s

Final Result

The initial speed of the body at point A is approximately 4.36 m/s. This calculation illustrates how the work done by an external agent, along with changes in potential energy, influences the kinetic energy and speed of an object in motion.