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.