A 0.34 kg particle has a speed of 2.0 m/s at point A and kinetic energy of 7.5 J at point B.
(a) What is its kinetic energy at A?
_____ J
(b) What is its speed at point B?
_____ m/s
(c) What is the total work done on the particle as it moves from A to B?
_____ J thank you so much
I will be happy to critique your thinking.
To answer these questions, we need to use the equations relating kinetic energy, mass, and speed.
(a) The kinetic energy (KE) is given by the equation KE = (1/2)mv^2, where m is the mass and v is the velocity (speed). In this case, we are given the mass of the particle (0.34 kg) and the speed at point A (2.0 m/s). Plugging these values into the equation, we can find the kinetic energy at point A:
KE_A = (1/2)(0.34 kg)(2.0 m/s)^2
Simplifying this equation, we find:
KE_A = (1/2)(0.34 kg)(4 m^2/s^2)
KE_A = 0.34 kg * 2 m^2/s^2
KE_A = 0.68 J
Therefore, the kinetic energy at point A is 0.68 J.
(b) To find the speed at point B, we can rearrange the kinetic energy equation and solve for the speed:
KE_B = (1/2)mv_B^2
Given that the kinetic energy at point B is 7.5 J and the mass is still 0.34 kg, we have:
7.5 J = (1/2)(0.34 kg)(v_B^2)
Now, solve for v_B:
v_B^2 = (7.5 J) / [(1/2)(0.34 kg)]
v_B^2 = (7.5 J) / (0.17 kg)
v_B^2 = 44.118 m^2/s^2
Taking the square root of both sides, we find:
v_B ≈ √(44.118 m^2/s^2)
v_B ≈ 6.64 m/s (rounded to two decimal places)
Therefore, the speed at point B is approximately 6.64 m/s.
(c) To find the total work done on the particle as it moves from A to B, we can use the work-energy theorem. The work (W) done on an object is equal to the change in its kinetic energy. The equation is given by:
W = KE_B - KE_A
Substituting the values we know, we have:
W = 7.5 J - 0.68 J
W ≈ 6.82 J (rounded to two decimal places)
Therefore, the total work done on the particle as it moves from A to B is approximately 6.82 J.