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A 0.15 kg baseball is moving 50 m/s. What is its kinetic energy?

11 answers

The kinetic energy (KE) of an object can be calculated using the formula:

\[ \text{KE} = \frac{1}{2} mv^2 \]

where \( m \) is the mass of the object in kilograms and \( v \) is the velocity in meters per second.

For a baseball with a mass of 0.15 kg moving at a velocity of 50 m/s, we can substitute the values into the formula:

\[ \text{KE} = \frac{1}{2} \times 0.15 , \text{kg} \times (50 , \text{m/s})^2 \]

Calculating this step by step:

  1. Calculate \( (50 , \text{m/s})^2 \):

\[ (50 , \text{m/s})^2 = 2500 , \text{m}^2/\text{s}^2 \]

  1. Now substitute back into the kinetic energy formula:

\[ \text{KE} = \frac{1}{2} \times 0.15 \times 2500 \]

  1. Calculate \( 0.15 \times 2500 \):

\[ 0.15 \times 2500 = 375 \]

  1. Now, compute \( \frac{1}{2} \times 375 \):

\[ \text{KE} = \frac{375}{2} = 187.5 , \text{J} \]

Thus, the kinetic energy of the baseball is \( 187.5 , \text{J} \).

A crane suspends a 1000kg pallet of steel 100 m in the air. g = 9.81 m/s/s.

To find the gravitational potential energy (PE) of the suspended pallet of steel, we can use the formula:

\[ \text{PE} = mgh \]

where:

  • \( m \) is the mass of the object (in kilograms),
  • \( g \) is the acceleration due to gravity (in meters per second squared),
  • \( h \) is the height above the reference point (in meters).

In this case:

  • \( m = 1000 , \text{kg} \) (mass of the pallet),
  • \( g = 9.81 , \text{m/s}^2 \) (acceleration due to gravity),
  • \( h = 100 , \text{m} \) (height above ground).

Now we can substitute these values into the potential energy formula:

\[ \text{PE} = 1000 , \text{kg} \times 9.81 , \text{m/s}^2 \times 100 , \text{m} \]

Calculating this step by step:

  1. Calculate \( m \times g \):

\[ 1000 , \text{kg} \times 9.81 , \text{m/s}^2 = 9810 , \text{N} \]

  1. Now, calculate the potential energy:

\[ \text{PE} = 9810 , \text{N} \times 100 , \text{m} = 981000 , \text{J} \]

Therefore, the gravitational potential energy of the suspended pallet of steel is \( 981000 , \text{J} \) or \( 981 , \text{kJ} \).

Question A large spring, has a spring constant of 10 N/m. It is stretched 2 m. What is the spring p

To find the potential energy stored in a spring that is stretched (or compressed), you can use the formula for spring potential energy (also known as elastic potential energy):

\[ PE = \frac{1}{2} k x^2 \]

where:

  • \( PE \) is the potential energy,
  • \( k \) is the spring constant (in N/m),
  • \( x \) is the displacement from the equilibrium position (in meters).

Given:

  • \( k = 10 , \text{N/m} \)
  • \( x = 2 , \text{m} \)

Now, substituting the values into the formula:

\[ PE = \frac{1}{2} \times 10 , \text{N/m} \times (2 , \text{m})^2 \]

Calculating it step by step:

  1. Calculate \( (2 , \text{m})^2 \):

\[ (2 , \text{m})^2 = 4 , \text{m}^2 \]

  1. Now substitute this back into the potential energy formula:

\[ PE = \frac{1}{2} \times 10 , \text{N/m} \times 4 , \text{m}^2 \]

  1. Calculate \( 10 , \text{N/m} \times 4 , \text{m}^2 \):

\[ 10 \times 4 = 40 \]

  1. Now calculate \( \frac{1}{2} \times 40 \):

\[ PE = \frac{40}{2} = 20 , \text{J} \]

Therefore, the potential energy stored in the spring when it is stretched 2 meters is 20 Joules.

What is the kinetic energy of a 2 kg glider moving at 3 m/s?

To find the kinetic energy (KE) of a glider, you can use the kinetic energy formula:

\[ \text{KE} = \frac{1}{2} mv^2 \]

where:

  • \( m \) is the mass of the object (in kilograms),
  • \( v \) is the velocity of the object (in meters per second).

Given:

  • \( m = 2 , \text{kg} \)
  • \( v = 3 , \text{m/s} \)

We can substitute these values into the formula:

\[ \text{KE} = \frac{1}{2} \times 2 , \text{kg} \times (3 , \text{m/s})^2 \]

Calculating it step by step:

  1. Calculate \( (3 , \text{m/s})^2 \):

\[ (3 , \text{m/s})^2 = 9 , \text{m}^2/\text{s}^2 \]

  1. Now substitute this back into the kinetic energy formula:

\[ \text{KE} = \frac{1}{2} \times 2 , \text{kg} \times 9 , \text{m}^2/\text{s}^2 \]

  1. Calculate \( 2 , \text{kg} \times 9 , \text{m}^2/\text{s}^2 \):

\[ 2 \times 9 = 18 \]

  1. Now calculate \( \frac{1}{2} \times 18 \):

\[ \text{KE} = \frac{18}{2} = 9 , \text{J} \]

Therefore, the kinetic energy of the 2 kg glider moving at 3 m/s is 9 Joules.

On planet easy math where the value of "g" is 10 m/s/s, there is a 2 kg model train is at the top of a 5 meter tall ramp. How much gravity potential energy does the model train have?

To calculate the gravitational potential energy (PE) of the model train at the top of the ramp, you can use the formula:

\[ PE = mgh \]

where:

  • \( PE \) is the gravitational potential energy (in joules),
  • \( m \) is the mass of the object (in kilograms),
  • \( g \) is the acceleration due to gravity (in meters per second squared),
  • \( h \) is the height above the reference point (in meters).

Given:

  • \( m = 2 , \text{kg} \)
  • \( g = 10 , \text{m/s}^2 \)
  • \( h = 5 , \text{m} \)

Now, substituting these values into the formula:

\[ PE = 2 , \text{kg} \times 10 , \text{m/s}^2 \times 5 , \text{m} \]

Calculating it step by step:

  1. Calculate \( 10 , \text{m/s}^2 \times 5 , \text{m} \):

\[ 10 \times 5 = 50 , \text{N} \]

  1. Now calculate \( 2 , \text{kg} \times 50 , \text{N} \):

\[ 2 , \text{kg} \times 50 , \text{N} = 100 , \text{J} \]

Therefore, the gravitational potential energy of the model train at the top of the 5-meter ramp is 100 Joules.

How much energy does a stiff spring with a spring constant of 10 N/m stretched 1 meter?

To calculate the potential energy stored in a spring that is stretched (or compressed), you can use the formula for spring potential energy (also known as elastic potential energy):

\[ PE = \frac{1}{2} k x^2 \]

where:

  • \( PE \) is the potential energy,
  • \( k \) is the spring constant (in N/m),
  • \( x \) is the displacement from the equilibrium position (in meters).

Given:

  • \( k = 10 , \text{N/m} \)
  • \( x = 1 , \text{m} \)

Now substituting these values into the formula:

\[ PE = \frac{1}{2} \times 10 , \text{N/m} \times (1 , \text{m})^2 \]

Calculating it step by step:

  1. Calculate \( (1 , \text{m})^2 \):

\[ (1 , \text{m})^2 = 1 , \text{m}^2 \]

  1. Substitute this back into the potential energy formula:

\[ PE = \frac{1}{2} \times 10 , \text{N/m} \times 1 , \text{m}^2 \]

  1. Calculate \( 10 , \text{N/m} \times 1 , \text{m}^2 \):

\[ 10 \times 1 = 10 \]

  1. Now calculate \( \frac{1}{2} \times 10 \):

\[ PE = \frac{10}{2} = 5 , \text{J} \]

Therefore, the energy stored in the stiff spring when it is stretched 1 meter is 5 Joules.