Two carts, one twice as heavy as the other, are at rest on a horizontal frictionless track. A person pushes each cart with the same force for 6.00 s.

If the kinetic energy of the lighter cart after the push is K, the kinetic energy of the heavier cart is

A. 1/4 K
B. 1/2 K
C. K
D. 2K
E. 4K

1/2 K

To solve this problem, we can use the concept of work and the work-energy theorem. The work done on an object is equal to the change in its kinetic energy.

Let's denote the force with which the person pushes the carts as F, and the mass of the lighter cart as m and the mass of the heavier cart as 2m.

The work done on each cart is given by the equation:

Work = Force * Distance

Since the carts are at rest, the distance is zero, so the work done on both carts is zero.

According to the work-energy theorem, the work done on an object is equal to the change in its kinetic energy. Therefore, the change in kinetic energy of both carts is zero.

Let's calculate the initial kinetic energy of the lighter cart. The kinetic energy formula is given by:

Kinetic energy = (1/2) * mass * velocity^2

Since the carts are at rest, the initial velocity of both carts is zero.

Kinetic energy of lighter cart = (1/2) * m * 0^2 = 0

Since the change in kinetic energy is zero, the final kinetic energy of the lighter cart is also zero.

So, the answer is option C. The kinetic energy of the heavier cart is K.

To solve this problem, we need to understand the concept of work and kinetic energy.

The work done on an object is defined as the product of the force applied on the object and the displacement of the object in the direction of the force. Mathematically, work (W) is given by the equation:

W = F * d * cos(theta)

In this case, since the force and displacement are in the same direction (horizontal), the angle between them (theta) is 0 degrees, so the cosine of theta is 1. Hence, the equation can be simplified to:

W = F * d

The work done on an object is equal to the change in its kinetic energy. Mathematically, the work-energy theorem states that:

W = delta(K)

Where W is the work done on the object and delta(K) is the change in kinetic energy.

In this problem, we are given that the carts are pushed with the same force for the same amount of time. This means that the work done on each cart will be the same.

The kinetic energy of an object is given by the equation:

K = 1/2 * m * v^2

Where K is the kinetic energy, m is the mass of the object, and v is the velocity of the object.

Let's consider the lighter cart first. Since the work done on the lighter cart is the same as the work done on the heavier cart, we can equate the work done on the lighter cart to the change in its kinetic energy:

W = delta(K_lighter)

Since the carts are initially at rest, the initial kinetic energy of the lighter cart is zero:

K_initial_lighter = 0

After the person pushes the lighter cart, it gains a velocity v_lighter. The final kinetic energy of the lighter cart can be calculated using the equation for kinetic energy:

K_final_lighter = 1/2 * m_lighter * v_lighter^2

Since we are given that the kinetic energy of the lighter cart after the push is K, we have:

K = K_final_lighter = 1/2 * m_lighter * v_lighter^2

Now, let's consider the heavier cart. Since the work done on the heavier cart is the same as the work done on the lighter cart, we can equate the work done on the heavier cart to the change in its kinetic energy:

W = delta(K_heavier)

Similarly, the initial kinetic energy of the heavier cart is zero:

K_initial_heavier = 0

After the person pushes the heavier cart, it gains a velocity v_heavier. The final kinetic energy of the heavier cart can be calculated using the equation for kinetic energy:

K_final_heavier = 1/2 * m_heavier * v_heavier^2

We are asked to find the ratio of the kinetic energy of the heavier cart to the kinetic energy of the lighter cart. Let's call this ratio R:

R = K_final_heavier / K_final_lighter

Substituting the values we have:

R = (1/2 * m_heavier * v_heavier^2) / (1/2 * m_lighter * v_lighter^2)

Since the masses of the carts are given in the problem as one being twice as heavy as the other:

m_heavier = 2 * m_lighter

Substituting this into the equation for R:

R = (1/2 * (2 * m_lighter) * v_heavier^2) / (1/2 * m_lighter * v_lighter^2)

R = (2 * m_lighter * v_heavier^2) / (m_lighter * v_lighter^2)

The masses of the lighters cancel out:

R = 2 * v_heavier^2 / v_lighter^2

However, the velocities of the carts are not given in the problem. We cannot determine the ratio R without knowing the velocities v_heavier and v_lighter. Therefore, the answer cannot be determined with the given information.

2K