A 0.568 kg particle has a speed of 1.57 m/s at point A and kinetic energy of 7.67 J at point B. What is its kinetic energy at A?

What is its speed at point B?

What is the total work done on the particle as it moves from A to B?

To find the kinetic energy at point A, we can use the formula for kinetic energy:

Kinetic Energy (KE) = 0.5 * mass * velocity^2

Given that the mass (m) is 0.568 kg and the speed (v) at point A is 1.57 m/s, we can substitute these values into the formula:

KE_A = 0.5 * 0.568 kg * (1.57 m/s)^2

Now we can calculate the kinetic energy at point A:

KE_A = 0.5 * 0.568 kg * (1.57 m/s)^2 = 0.443 J

Therefore, the kinetic energy at point A is 0.443 J.

To find the speed at point B, we can use the formula for kinetic energy and rearrange it:

Kinetic Energy (KE) = 0.5 * mass * velocity^2

Given that the kinetic energy at point B is 7.67 J and the mass is 0.568 kg, we can rearrange the formula to solve for velocity (v):

KE_B = 0.5 * 0.568 kg * v^2

Now we can solve for velocity (v):

v^2 = (2 * KE_B) / m

v = sqrt((2 * KE_B) / m) = sqrt((2 * 7.67 J) / 0.568 kg) ≈ 3.96 m/s

Therefore, the speed at point B is approximately 3.96 m/s.

To find the total work done on the particle as it moves from point A to point B, we can use the work-energy theorem:

Work (W) = Change in Kinetic Energy (ΔKE)

Given that the initial kinetic energy at point A is 0.443 J and the final kinetic energy at point B is 7.67 J, we can calculate the total work done:

W = ΔKE = KE_B - KE_A = 7.67 J - 0.443 J ≈ 7.23 J

Therefore, the total work done on the particle as it moves from point A to point B is approximately 7.23 J.