Sure, let’s solve each problem step-by-step.
1. Kinetic Energy of the Ball
The formula for kinetic energy (KE) is given by:
\[ KE = \frac{1}{2} mv^2 \]
Where:
- \(m\) = mass of the object (in kg)
- \(v\) = velocity of the object (in m/s)
For the ball:
- \(m = 2 , \text{kg}\)
- \(v = 10 , \text{m/s}\)
Substituting the values:
\[ KE = \frac{1}{2} \times 2 , \text{kg} \times (10 , \text{m/s})^2 \] \[ KE = \frac{1}{2} \times 2 , \times 100 \] \[ KE = 1 \times 100 = 100 , \text{J} \]
The kinetic energy possessed by the ball is 100 Joules.
2. Kinetic Energy of the Spaceship
Using the same formula for kinetic energy:
For the spaceship:
- \(m = 20000 , \text{kg}\)
- \(v = 10 , \text{m/s}\)
Substituting the values:
\[ KE = \frac{1}{2} \times 20000 , \text{kg} \times (10 , \text{m/s})^2 \] \[ KE = \frac{1}{2} \times 20000 , \times 100 \] \[ KE = 10000 \times 100 = 1000000 , \text{J} \]
The kinetic energy possessed by the spaceship is 1,000,000 Joules.
3. New Speed After Work Done
The work-energy principle states that the work done on an object is equal to the change in kinetic energy.
Given:
- Work done \(W = -50 , \text{J}\)
- Initial speed \(v_i = 10 , \text{m/s}\)
- Mass \(m = 2 , \text{kg}\)
First, find the initial kinetic energy:
\[ KE_i = \frac{1}{2} mv_i^2 = \frac{1}{2} \times 2 \times (10)^2 = 100 , \text{J} \]
Now, calculate the final kinetic energy:
\[ KE_f = KE_i + W = 100 , \text{J} - 50 , \text{J} = 50 , \text{J} \]
Using the final kinetic energy to find the final speed \(v_f\):
\[ KE_f = \frac{1}{2} mv_f^2 \] \[ 50 = \frac{1}{2} \times 2 \times v_f^2 \] \[ 50 = v_f^2 \] \[ v_f = \sqrt{50} \approx 7.07 , \text{m/s} \]
The new speed of the object is approximately 7.07 m/s.
4. Potential Energy Just Before Hitting the Ground
Potential energy (PE) when at a height \(h\) is given by:
\[ PE = mgh \]
where:
- \(m\) = mass,
- \(g\) = acceleration due to gravity (approximately \(9.81 , \text{m/s}^2\)),
- \(h\) = height above ground (assume \(h = 0\) just before hitting the ground).
If you are just about to hit the ground, your height \(h = 0\):
\[ PE = m \cdot g \cdot 0 = 0 , \text{J} \]
The potential energy just before hitting the ground is 0 Joules.
5. Potential Energy of the Block on the Wedge
For the mass taken uphill, the potential energy can also be calculated as:
\[ PE = mgh \]
First, we need to calculate the height \(h\) using the wedge inclination. The height can be found using the sine function since we have the angle \(\theta\) and the vertical displacement will be:
\[ h = d \cdot \sin(\theta) \]
Where:
- \(d = 10 , \text{m}\)
- \(\theta = 30^\circ\)
Calculating height \(h\):
\[ h = 10 \cdot \sin(30^\circ) = 10 \cdot 0.5 = 5 , \text{m} \]
Now calculate the potential energy:
\[ PE = mgh = 10 , \text{kg} \cdot 9.81 , \text{m/s}^2 \cdot 5 , \text{m} \] \[ PE = 10 \cdot 9.81 \cdot 5 = 490.5 , \text{J} \]
The potential energy of the block is approximately 490.5 Joules.
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