1.) At one point in its fall, an object (m=2.5 kg.) has a kinetic energy of 5 J(joule) and a potential energy of 10 J.
a) what is the total energy of the system?
b) what is the velocity at this point?
The object continues to another point where its potential energy is 3J(joule).
c) what is the KE(kinetic energy) at this point.
d) what is the velocity at this point.
e)how high off of the ground is the object.
>> FORMULAS <<
KE=1/2 v^2
PE= mgh
_please help me! >.<
a) E(total) =KE+PE=5+10 =15 J
b) v=sqrt(2•KE/m)=…(m/s)
c) KE1=15 -3 =12 J
d) v1=sqrt(2•KE1/m)=…(m/s)
e) PE= mgh = 3J
h=3/mg=3/2.5•9.8=.... (m)
Sure! Let's solve the problems step-by-step:
a) To find the total energy of the system, we need to sum up the kinetic energy (KE) and potential energy (PE). Mathematically, it is given by the equation:
Total energy = KE + PE
Given that the kinetic energy is 5 J and the potential energy is 10 J, we can substitute these values into the equation:
Total energy = 5 J + 10 J
Total energy = 15 J
Therefore, the total energy of the system is 15 J.
b) To find the velocity at this point, we can use the equation for kinetic energy:
KE = 1/2 * m * v^2
Given that the kinetic energy is 5 J and the mass (m) is 2.5 kg, we can rearrange the equation and solve for the velocity (v):
v^2 = (2 * KE) / m
v^2 = (2 * 5 J) / (2.5 kg)
v^2 = 10 J / 2.5 kg
v^2 = 4 m^2/s^2
Taking the square root of both sides, we find:
v = 2 m/s
Therefore, the velocity at this point is 2 m/s.
c) To find the kinetic energy at the new point where the potential energy is 3 J, we can use the equation for potential energy to determine the change in potential energy:
ΔPE = PE_final - PE_initial
Given that the potential energy at the new point is 3 J and the initial potential energy is 10 J, we can substitute these values into the equation:
ΔPE = 3 J - 10 J
ΔPE = -7 J
Since kinetic energy and potential energy are interconvertible, the change in potential energy will result in a corresponding change in kinetic energy. Therefore, the kinetic energy at this point will also be -7 J.
d) To find the velocity at this point, we can use the equation for kinetic energy:
KE = 1/2 * m * v^2
Given that the kinetic energy is -7 J and the mass (m) is still 2.5 kg, we can rearrange the equation and solve for the velocity (v):
v^2 = (2 * KE) / m
v^2 = (2 * (-7 J)) / (2.5 kg)
v^2 = -14 J / 2.5 kg
v^2 = -5.6 m^2/s^2
However, the negative square root does not make physical sense in this scenario. Therefore, the velocity at this point is not defined.
e) To determine how high off the ground the object is, we can use the equation for potential energy:
PE = m * g * h
Given that the potential energy is 3 J, the mass (m) is 2.5 kg, and the gravitational acceleration (g) is typically 9.8 m/s^2, we can rearrange the equation and solve for the height (h):
h = PE / (m * g)
h = 3 J / (2.5 kg * 9.8 m/s^2)
h ≈ 0.122 m
Therefore, the object is approximately 0.122 meters high off the ground.
Sure! Let's go through each of the questions step by step using the given formulas.
a) To find the total energy of the system, we need to add the kinetic energy and potential energy together. The formula for total energy (E) is:
E = KE + PE
Substituting the given values:
E = 5 J + 10 J
We get:
E = 15 J
Therefore, the total energy of the system is 15 Joules.
b) To find the velocity at this point, we need to use the formula for kinetic energy:
KE = 1/2 * m * v^2
Rearranging the formula to solve for velocity (v), we get:
v = √(2 * KE / m)
Substituting the given values:
v = √(2 * 5 J / 2.5 kg)
Simplifying:
v = √(2) m/s
Therefore, the velocity at this point is √2 m/s.
c) To find the kinetic energy at the second point, we can use the given potential energy and the total energy (which remains constant throughout the fall) using the equation:
E = KE + PE
Substituting the given values:
15 J = KE + 3 J
Simplifying:
KE = 15 J - 3 J
KE = 12 J
Therefore, the kinetic energy at this point is 12 Joules.
d) Using the formula for kinetic energy as before, we can find the velocity at this second point:
KE = 1/2 * m * v^2
Rearranging the formula to solve for velocity (v), we get:
v = √(2 * KE / m)
Substituting the given values:
v = √(2 * 12 J / 2.5 kg)
Simplifying:
v = √(48/5) m/s
Therefore, the velocity at this point is √(48/5) m/s.
e) To find the height off the ground at the second point, we can use the formula for potential energy:
PE = m * g * h
Rearranging the formula to solve for height (h), we get:
h = PE / (m * g)
Substituting the given values:
h = 3 J / (2.5 kg * g)
Since the value of 'g' (acceleration due to gravity) is approximated as 9.8 m/s^2:
h = 3 J / (2.5 kg * 9.8 m/s^2)
Simplifying:
h ≈ 0.122 m
Therefore, the object is approximately 0.122 meters high off the ground at the second point.
I hope this helps! Let me know if you have any further questions.