Question
Assume the toy car with mass 150 g starts at rest and there is no friction.
Calculate The Total Mechanical Energy, Kinetic Energy, Gravitational Potential Energy, and
velocity at following positions:
1. Speed = 0.0cm/s
Time = 70s
2. Speed = 160.0cm/s
Time = 20s
3. Speed = 280.0cm/s
Time = 20s
4. Speed = 400.0cm/s
Time = 60s
5. Speed = 440.0cm/s
Time = 90s
Using the formulas below, solve the answers above.
Eg = (m)(g)(△h)
Ek = 1/2 mv^2
v = √((2Ek)/(m))
Et = Ek + Eg
Calculate The Total Mechanical Energy, Kinetic Energy, Gravitational Potential Energy, and
velocity at following positions:
1. Speed = 0.0cm/s
Time = 70s
2. Speed = 160.0cm/s
Time = 20s
3. Speed = 280.0cm/s
Time = 20s
4. Speed = 400.0cm/s
Time = 60s
5. Speed = 440.0cm/s
Time = 90s
Using the formulas below, solve the answers above.
Eg = (m)(g)(△h)
Ek = 1/2 mv^2
v = √((2Ek)/(m))
Et = Ek + Eg
Answers
Answered by
GPT 3.5
To calculate the total mechanical energy (Et), kinetic energy (Ek), gravitational potential energy (Eg), and velocity (v) at each given position, we can use the formulas provided.
Given:
Mass (m) = 150 g = 0.15 kg
Acceleration due to gravity (g) = 9.8 m/s^2
1. At Speed = 0.0 cm/s, Time = 70s:
Since the speed is 0, the kinetic energy (Ek) is also 0.
The potential energy (Eg) at this position can be calculated using the formula:
Eg = (m)(g)(Δh)
Eg = (0.15 kg)(9.8 m/s^2)(0) = 0 J
Therefore, the total mechanical energy (Et) is also 0.
The velocity (v) can be calculated using the formula:
v = √((2Ek)/(m))
v = √((2(0))/(0.15 kg)) = 0 m/s
2. At Speed = 160.0 cm/s, Time = 20s:
The velocity (v) is given as 160 cm/s. Converting it to m/s, we get 1.6 m/s.
The kinetic energy (Ek) can be calculated using the formula:
Ek = 1/2 mv^2
Ek = 1/2 (0.15 kg)(1.6 m/s)^2 ≈ 0.192 J
The potential energy (Eg) at this position is 0, as the height (Δh) is not given.
Therefore, the total mechanical energy (Et) is approximately equal to the kinetic energy:
Et ≈ 0.192 J
The velocity (v) is already provided as 1.6 m/s.
3. At Speed = 280.0 cm/s, Time = 20s:
The velocity (v) is given as 280 cm/s. Converting it to m/s, we get 2.8 m/s.
The kinetic energy (Ek) can be calculated using the formula:
Ek = 1/2 mv^2
Ek = 1/2 (0.15 kg)(2.8 m/s)^2 ≈ 0.588 J
The potential energy (Eg) at this position is 0, as the height (Δh) is not given.
Therefore, the total mechanical energy (Et) is approximately equal to the kinetic energy:
Et ≈ 0.588 J
The velocity (v) is already provided as 2.8 m/s.
4. At Speed = 400.0 cm/s, Time = 60s:
The velocity (v) is given as 400 cm/s. Converting it to m/s, we get 4 m/s.
The kinetic energy (Ek) can be calculated using the formula:
Ek = 1/2 mv^2
Ek = 1/2 (0.15 kg)(4 m/s)^2 = 0.48 J
The potential energy (Eg) at this position is 0, as the height (Δh) is not given.
Therefore, the total mechanical energy (Et) is approximately equal to the kinetic energy:
Et = 0.48 J
The velocity (v) is already provided as 4 m/s.
5. At Speed = 440.0 cm/s, Time = 90s:
The velocity (v) is given as 440 cm/s. Converting it to m/s, we get 4.4 m/s.
The kinetic energy (Ek) can be calculated using the formula:
Ek = 1/2 mv^2
Ek = 1/2 (0.15 kg)(4.4 m/s)^2 ≈ 0.726 J
The potential energy (Eg) at this position is 0, as the height (Δh) is not given.
Therefore, the total mechanical energy (Et) is approximately equal to the kinetic energy:
Et ≈ 0.726 J
The velocity (v) is already provided as 4.4 m/s.
In summary:
1. Total Mechanical Energy (Et) = 0 J, Kinetic Energy (Ek) = 0 J, Gravitational Potential Energy (Eg) = 0 J, Velocity (v) = 0 m/s.
2. Total Mechanical Energy (Et) ≈ 0.192 J, Kinetic Energy (Ek) ≈ 0.192 J, Gravitational Potential Energy (Eg) = 0 J, Velocity (v) = 1.6 m/s.
3. Total Mechanical Energy (Et) ≈ 0.588 J, Kinetic Energy (Ek) ≈ 0.588 J, Gravitational Potential Energy (Eg) = 0 J, Velocity (v) = 2.8 m/s.
4. Total Mechanical Energy (Et) = 0.48 J, Kinetic Energy (Ek) = 0.48 J, Gravitational Potential Energy (Eg) = 0 J, Velocity (v) = 4 m/s.
5. Total Mechanical Energy (Et) ≈ 0.726 J, Kinetic Energy (Ek) ≈ 0.726 J, Gravitational Potential Energy (Eg) = 0 J, Velocity (v) = 4.4 m/s.
Given:
Mass (m) = 150 g = 0.15 kg
Acceleration due to gravity (g) = 9.8 m/s^2
1. At Speed = 0.0 cm/s, Time = 70s:
Since the speed is 0, the kinetic energy (Ek) is also 0.
The potential energy (Eg) at this position can be calculated using the formula:
Eg = (m)(g)(Δh)
Eg = (0.15 kg)(9.8 m/s^2)(0) = 0 J
Therefore, the total mechanical energy (Et) is also 0.
The velocity (v) can be calculated using the formula:
v = √((2Ek)/(m))
v = √((2(0))/(0.15 kg)) = 0 m/s
2. At Speed = 160.0 cm/s, Time = 20s:
The velocity (v) is given as 160 cm/s. Converting it to m/s, we get 1.6 m/s.
The kinetic energy (Ek) can be calculated using the formula:
Ek = 1/2 mv^2
Ek = 1/2 (0.15 kg)(1.6 m/s)^2 ≈ 0.192 J
The potential energy (Eg) at this position is 0, as the height (Δh) is not given.
Therefore, the total mechanical energy (Et) is approximately equal to the kinetic energy:
Et ≈ 0.192 J
The velocity (v) is already provided as 1.6 m/s.
3. At Speed = 280.0 cm/s, Time = 20s:
The velocity (v) is given as 280 cm/s. Converting it to m/s, we get 2.8 m/s.
The kinetic energy (Ek) can be calculated using the formula:
Ek = 1/2 mv^2
Ek = 1/2 (0.15 kg)(2.8 m/s)^2 ≈ 0.588 J
The potential energy (Eg) at this position is 0, as the height (Δh) is not given.
Therefore, the total mechanical energy (Et) is approximately equal to the kinetic energy:
Et ≈ 0.588 J
The velocity (v) is already provided as 2.8 m/s.
4. At Speed = 400.0 cm/s, Time = 60s:
The velocity (v) is given as 400 cm/s. Converting it to m/s, we get 4 m/s.
The kinetic energy (Ek) can be calculated using the formula:
Ek = 1/2 mv^2
Ek = 1/2 (0.15 kg)(4 m/s)^2 = 0.48 J
The potential energy (Eg) at this position is 0, as the height (Δh) is not given.
Therefore, the total mechanical energy (Et) is approximately equal to the kinetic energy:
Et = 0.48 J
The velocity (v) is already provided as 4 m/s.
5. At Speed = 440.0 cm/s, Time = 90s:
The velocity (v) is given as 440 cm/s. Converting it to m/s, we get 4.4 m/s.
The kinetic energy (Ek) can be calculated using the formula:
Ek = 1/2 mv^2
Ek = 1/2 (0.15 kg)(4.4 m/s)^2 ≈ 0.726 J
The potential energy (Eg) at this position is 0, as the height (Δh) is not given.
Therefore, the total mechanical energy (Et) is approximately equal to the kinetic energy:
Et ≈ 0.726 J
The velocity (v) is already provided as 4.4 m/s.
In summary:
1. Total Mechanical Energy (Et) = 0 J, Kinetic Energy (Ek) = 0 J, Gravitational Potential Energy (Eg) = 0 J, Velocity (v) = 0 m/s.
2. Total Mechanical Energy (Et) ≈ 0.192 J, Kinetic Energy (Ek) ≈ 0.192 J, Gravitational Potential Energy (Eg) = 0 J, Velocity (v) = 1.6 m/s.
3. Total Mechanical Energy (Et) ≈ 0.588 J, Kinetic Energy (Ek) ≈ 0.588 J, Gravitational Potential Energy (Eg) = 0 J, Velocity (v) = 2.8 m/s.
4. Total Mechanical Energy (Et) = 0.48 J, Kinetic Energy (Ek) = 0.48 J, Gravitational Potential Energy (Eg) = 0 J, Velocity (v) = 4 m/s.
5. Total Mechanical Energy (Et) ≈ 0.726 J, Kinetic Energy (Ek) ≈ 0.726 J, Gravitational Potential Energy (Eg) = 0 J, Velocity (v) = 4.4 m/s.