How Does the Law of Conservation of Energy Apply to a Particle's Motion?

[dot123]

Homework Statement

A particle of mass 0.400 kg is shot from P . The particle has an initial velocity vi with the horizontal component of 30 m/s. The particle rises to a maximum height of 20 m above P. Using the law of conservation of energy, determine :
(a) the vertical component of vi
(b) the work done by the gravitational force on the particle during its motion from P to B
(c) the horizontal and the vertical components of the velocity vector when the particle reaches B
P is in height of 60 m.

Homework Equations

Ek = -Ep

The Attempt at a Solution

a.)
1/2mv2 = -(m*g*h)
1/2v2 = -(-9,8*20)
v2 = 2*9,8m/s2*20m
v=19,7 m/s¨
b.)
A=m*g*h
A=0,4kg*9,8m/s2*60m
A=294 J
c.)
1/2mv2 =-(m*g*h)
1/2v2=-(9,8*-80)
v2=2*9,8m/s2*80m
v=39,5m/s

I do not know if I should calculate with minus Ep. And then if the acceleration should be with minus in a.) and in c.) the acceleration positive and height negative?

 

[barryj]

I am a bit confused. You say the particle reaches a max heighth of 20 m but use use 60 m in the solution. Seems like part a is correct. The first statement of part b is correct but the second line makes no sense to me.

 

[dot123]

Sorry, it is confusing. The particle is shot from P (P is in the height of 60m above the ground) and then particle rises to height of 20m above P so the particle is in the total height of 80m above the ground and then reaches the ground - B . I hope that I made it clear.

 

[haruspex]

You can choose to measure distance upwards or downwards - entirely up to you, so long as you are consistent. Note that swapping swaps the sign on both acceleration and height, so leaves the sign of PE unchanged.

 

[Nickie]

I would like to clarify and provide a more detailed explanation for the solution to this problem using the law of conservation of energy.

The law of conservation of energy states that energy cannot be created or destroyed, but can only be transferred or transformed from one form to another. In this problem, we can see that the particle has both kinetic energy (Ek) and potential energy (Ep) at different points in its motion.

(a) To determine the vertical component of the initial velocity (vi), we can use the conservation of energy equation: Ek = -Ep. This means that the initial kinetic energy of the particle (1/2mv^2) is equal to the negative potential energy at point P (mgh). We can rearrange this equation to solve for vi, and we get:

1/2mv^2 = -mgh
v^2 = -2gh
v = √(-2gh)

Since we know the mass (m = 0.400 kg) and the height (h = 20 m) from the given information, we can plug in these values and solve for the initial velocity vi. This gives us a result of vi = -19.8 m/s, which is the negative of the actual vertical component of the initial velocity. This is because the particle is initially moving upwards, so the vertical component of the velocity should be positive. Therefore, the correct answer for part (a) is vi = 19.8 m/s.

(b) The work done by the gravitational force on the particle during its motion from P to B can also be calculated using the conservation of energy equation: Ek = -Ep. In this case, we are interested in the change in potential energy (ΔEp) as the particle moves from point P to B. This can be calculated as:

ΔEp = -mghB + mghP
ΔEp = -mghB + mghP
ΔEp = -0.400 kg * 9.8 m/s^2 * (60 m - 20 m)
ΔEp = -0.400 kg * 9.8 m/s^2 * 40 m
ΔEp = -156.8 J

Since the change in potential energy is negative, this means that the gravitational force is doing work on the particle as it moves from P to B. The correct answer for part (b) is therefore -156.