A particle of mass 500 gm is moving in a straight line with velocity v = bx^5/2. The work done by the net force during its displacement from x = 0 to x = 4 m is : (Take b = 0.25 m^–3/2s^–1).

Question

Two solid spheres each of mass 2 kg and radius 75 cm are arranged as shown. Find MOI of the system about the given axis.

Answer 1

To determine the work done by the net force during the displacement of a particle from \(x = 0\) to \(x = 4\) m, we will follow these steps:

Firstly, identify the given information:
- Mass of the particle, \(m = 500 \, \text{gm} = 0.5 \, \text{kg}\)
- Velocity of the particle, \(v = bx^{5/2}\)
- Where \(b = 0.25 \, \text{m}^{-3/2}\text{s}^{-1}\)
- The particle moves from \(x = 0\) to \(x = 4\) m.
Use the relation for kinetic energy: \(K.E. = \frac{1}{2}mv^2\).
Velocity as a function of position is given as \(v = bx^{5/2}\). Thus, the kinetic energy as a function of \(x\) can be expressed as:
K.E.(x) = \frac{1}{2}m(bx^{5/2})^2 = \frac{1}{2}m b^2 x^5
Now, to find the change in kinetic energy (which is equal to the work done by the net force), calculate the kinetic energy at \(x = 0\) and \(x = 4\), then find the difference:
- At \(x = 0\), \(\text{K.E.}(0) = \frac{1}{2} \cdot 0.5 \cdot (0.25)^2 \cdot 0^5 = 0 \, \text{J}\)
- At \(x = 4\), \(\text{K.E.}(4) = \frac{1}{2} \cdot 0.5 \cdot (0.25)^2 \cdot 4^5 \)
- Calculating further, \(\text{K.E.}(4) = 0.125 \cdot 0.0625 \cdot 1024 = 16 \, \text{J}\)
Therefore, the work done by the net force is the change in kinetic energy: \Delta \text{K.E.} = \text{K.E.}(4) - \text{K.E.}(0) = 16 \, \text{J} - 0 \, \text{J} = 16 \, \text{J}
Hence, the correct answer is: 16 J.

Answer 2

To determine the work done by the net force during the displacement of a particle from \(x = 0\) to \(x = 4\) m, we will follow these steps:

Firstly, identify the given information:
- Mass of the particle, \(m = 500 \, \text{gm} = 0.5 \, \text{kg}\)
- Velocity of the particle, \(v = bx^{5/2}\)
- Where \(b = 0.25 \, \text{m}^{-3/2}\text{s}^{-1}\)
- The particle moves from \(x = 0\) to \(x = 4\) m.
Use the relation for kinetic energy: \(K.E. = \frac{1}{2}mv^2\).
Velocity as a function of position is given as \(v = bx^{5/2}\). Thus, the kinetic energy as a function of \(x\) can be expressed as:
K.E.(x) = \frac{1}{2}m(bx^{5/2})^2 = \frac{1}{2}m b^2 x^5
Now, to find the change in kinetic energy (which is equal to the work done by the net force), calculate the kinetic energy at \(x = 0\) and \(x = 4\), then find the difference:
- At \(x = 0\), \(\text{K.E.}(0) = \frac{1}{2} \cdot 0.5 \cdot (0.25)^2 \cdot 0^5 = 0 \, \text{J}\)
- At \(x = 4\), \(\text{K.E.}(4) = \frac{1}{2} \cdot 0.5 \cdot (0.25)^2 \cdot 4^5 \)
- Calculating further, \(\text{K.E.}(4) = 0.125 \cdot 0.0625 \cdot 1024 = 16 \, \text{J}\)
Therefore, the work done by the net force is the change in kinetic energy: \Delta \text{K.E.} = \text{K.E.}(4) - \text{K.E.}(0) = 16 \, \text{J} - 0 \, \text{J} = 16 \, \text{J}
Hence, the correct answer is: 16 J.