To solve the above problems, we will use the given formulas for calculating total mechanical energy (Et), kinetic energy (Ek), gravitational potential energy (Eg), and velocity (v). Here are the steps to solve each problem:
1. Speed = 0.0 cm/s, Time = 70 s
Since the speed is 0, the kinetic energy is 0. We can assume the toy car is at the bottom position. Therefore, the gravitational potential energy is maximum.
Ek = 0 J
Eg = (0.15 kg)(9.8 m/s^2)(0 m) = 0 J
Et = Ek + Eg = 0 J
v = √((2Ek)/(m)) = √((2*0)/(0.15)) = 0 m/s
2. Speed = 160.0 cm/s, Time = 20 s
We are given the speed, so we can calculate the kinetic energy using the formula.
Ek = 1/2 * (0.15 kg) * (160 cm/s)^2 = 192 J
Since the speed is not 0, the toy car is not at the bottom position. Therefore, the gravitational potential energy is less than maximum.
Eg = (0.15 kg)(9.8 m/s^2)(△h)
As we don't know the change in height (△h), we cannot calculate Eg.
Et = Ek + Eg = Unknown
v = √((2Ek)/(m)) = √((2*192)/(0.15)) = 57.95 m/s
3. Speed = 280.0 cm/s, Time = 20 s
Using the given speed, we can calculate the kinetic energy.
Ek = 1/2 * (0.15 kg) * (280 cm/s)^2 = 588 J
Since the speed is not 0, the toy car is not at the bottom position. Therefore, the gravitational potential energy is less than maximum.
Eg = (0.15 kg)(9.8 m/s^2)(△h)
As we don't know the change in height (△h), we cannot calculate Eg.
Et = Ek + Eg = Unknown
v = √((2Ek)/(m)) = √((2*588)/(0.15)) = 81.98 m/s
4. Speed = 400.0 cm/s, Time = 60 s
Using the given speed, we can calculate the kinetic energy.
Ek = 1/2 * (0.15 kg) * (400 cm/s)^2 = 1200 J
Since the speed is not 0, the toy car is not at the bottom position. Therefore, the gravitational potential energy is less than maximum.
Eg = (0.15 kg)(9.8 m/s^2)(△h)
As we don't know the change in height (△h), we cannot calculate Eg.
Et = Ek + Eg = Unknown
v = √((2Ek)/(m)) = √((2*1200)/(0.15)) = 122.47 m/s
5. Speed = 440.0 cm/s, Time = 90 s
Using the given speed, we can calculate the kinetic energy.
Ek = 1/2 * (0.15 kg) * (440 cm/s)^2 = 1452 J
Since the speed is not 0, the toy car is not at the bottom position. Therefore, the gravitational potential energy is less than maximum.
Eg = (0.15 kg)(9.8 m/s^2)(△h)
As we don't know the change in height (△h), we cannot calculate Eg.
Et = Ek + Eg = Unknown
v = √((2Ek)/(m)) = √((2*1452)/(0.15)) = 134.13 m/s
Please note that without the information about the change in height (△h), we cannot calculate the gravitational potential energy (Eg) and hence the total mechanical energy (Et).
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.
Et = Ek + Eg
Ek = 1/2 mv^2
Eg = (m)(g)(△h)
v = √((2Ek)/(m))
3 answers
are you sure these are answers are correct?
Apologies for the incorrect answers. Let's go through the calculations again to obtain the correct values.
1. Speed = 0.0 cm/s, Time = 70 s
Since the speed is 0, the kinetic energy is 0. The toy car is assumed to be at the bottom position, so the gravitational potential energy is maximum.
Ek = 0 J
Eg = (0.15 kg)(9.8 m/s^2)(0 m) = 0 J
Et = Ek + Eg = 0 J
v = √((2Ek)/(m)) = √((2*0)/(0.15)) = 0 m/s
2. Speed = 160.0 cm/s, Time = 20 s
Using the given speed, we can calculate the kinetic energy.
Ek = 1/2 * (0.15 kg) * (160 cm/s)^2 = 192 J
Since the speed is not 0, the toy car is not at the bottom position. Therefore, the gravitational potential energy is less than maximum.
Eg = (0.15 kg)(9.8 m/s^2)(△h)
As we don't know the change in height (△h), we cannot calculate Eg.
Et = Ek + Eg = Unknown
v = √((2Ek)/(m)) = √((2*192)/(0.15)) = 45.10 m/s
3. Speed = 280.0 cm/s, Time = 20 s
Using the given speed, we can calculate the kinetic energy.
Ek = 1/2 * (0.15 kg) * (280 cm/s)^2 = 588 J
Since the speed is not 0, the toy car is not at the bottom position. Therefore, the gravitational potential energy is less than maximum.
Eg = (0.15 kg)(9.8 m/s^2)(△h)
As we don't know the change in height (△h), we cannot calculate Eg.
Et = Ek + Eg = Unknown
v = √((2Ek)/(m)) = √((2*588)/(0.15)) = 68.35 m/s
4. Speed = 400.0 cm/s, Time = 60 s
Using the given speed, we can calculate the kinetic energy.
Ek = 1/2 * (0.15 kg) * (400 cm/s)^2 = 1200 J
Since the speed is not 0, the toy car is not at the bottom position. Therefore, the gravitational potential energy is less than maximum.
Eg = (0.15 kg)(9.8 m/s^2)(△h)
As we don't know the change in height (△h), we cannot calculate Eg.
Et = Ek + Eg = Unknown
v = √((2Ek)/(m)) = √((2*1200)/(0.15)) = 122.47 m/s
5. Speed = 440.0 cm/s, Time = 90 s
Using the given speed, we can calculate the kinetic energy.
Ek = 1/2 * (0.15 kg) * (440 cm/s)^2 = 1452 J
Since the speed is not 0, the toy car is not at the bottom position. Therefore, the gravitational potential energy is less than maximum.
Eg = (0.15 kg)(9.8 m/s^2)(△h)
As we don't know the change in height (△h), we cannot calculate Eg.
Et = Ek + Eg = Unknown
v = √((2Ek)/(m)) = √((2*1452)/(0.15)) = 134.13 m/s
Again, please note that without the information about the change in height (△h), we cannot calculate the gravitational potential energy (Eg) and hence the total mechanical energy (Et).
1. Speed = 0.0 cm/s, Time = 70 s
Since the speed is 0, the kinetic energy is 0. The toy car is assumed to be at the bottom position, so the gravitational potential energy is maximum.
Ek = 0 J
Eg = (0.15 kg)(9.8 m/s^2)(0 m) = 0 J
Et = Ek + Eg = 0 J
v = √((2Ek)/(m)) = √((2*0)/(0.15)) = 0 m/s
2. Speed = 160.0 cm/s, Time = 20 s
Using the given speed, we can calculate the kinetic energy.
Ek = 1/2 * (0.15 kg) * (160 cm/s)^2 = 192 J
Since the speed is not 0, the toy car is not at the bottom position. Therefore, the gravitational potential energy is less than maximum.
Eg = (0.15 kg)(9.8 m/s^2)(△h)
As we don't know the change in height (△h), we cannot calculate Eg.
Et = Ek + Eg = Unknown
v = √((2Ek)/(m)) = √((2*192)/(0.15)) = 45.10 m/s
3. Speed = 280.0 cm/s, Time = 20 s
Using the given speed, we can calculate the kinetic energy.
Ek = 1/2 * (0.15 kg) * (280 cm/s)^2 = 588 J
Since the speed is not 0, the toy car is not at the bottom position. Therefore, the gravitational potential energy is less than maximum.
Eg = (0.15 kg)(9.8 m/s^2)(△h)
As we don't know the change in height (△h), we cannot calculate Eg.
Et = Ek + Eg = Unknown
v = √((2Ek)/(m)) = √((2*588)/(0.15)) = 68.35 m/s
4. Speed = 400.0 cm/s, Time = 60 s
Using the given speed, we can calculate the kinetic energy.
Ek = 1/2 * (0.15 kg) * (400 cm/s)^2 = 1200 J
Since the speed is not 0, the toy car is not at the bottom position. Therefore, the gravitational potential energy is less than maximum.
Eg = (0.15 kg)(9.8 m/s^2)(△h)
As we don't know the change in height (△h), we cannot calculate Eg.
Et = Ek + Eg = Unknown
v = √((2Ek)/(m)) = √((2*1200)/(0.15)) = 122.47 m/s
5. Speed = 440.0 cm/s, Time = 90 s
Using the given speed, we can calculate the kinetic energy.
Ek = 1/2 * (0.15 kg) * (440 cm/s)^2 = 1452 J
Since the speed is not 0, the toy car is not at the bottom position. Therefore, the gravitational potential energy is less than maximum.
Eg = (0.15 kg)(9.8 m/s^2)(△h)
As we don't know the change in height (△h), we cannot calculate Eg.
Et = Ek + Eg = Unknown
v = √((2Ek)/(m)) = √((2*1452)/(0.15)) = 134.13 m/s
Again, please note that without the information about the change in height (△h), we cannot calculate the gravitational potential energy (Eg) and hence the total mechanical energy (Et).