A body of 4.0 kg is lying at rest. Under the action of a constant force, it gains a speed of 5 m/s. The work done by the force will be _______.
50J
Calculating Work Done by a Constant Force
The question asks us to find the work done by a constant force acting on a body that starts from rest and gains a specific speed. We are given the mass of the body and its initial and final speeds.
We can use the work-energy theorem to solve this problem. The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy.
Work Done ($W$) = Change in Kinetic Energy ($\Delta KE$)
Change in Kinetic Energy ($\Delta KE$) = Final Kinetic Energy ($KE_f$) - Initial Kinetic Energy ($KE_i$).
Initial and Final Kinetic Energy Calculation
The formula for kinetic energy is given by:
\(KE = \frac{1}{2}mv^2\)
where:
- \(m\) is the mass of the body
- \(v\) is the speed of the body
Initial Kinetic Energy
The body starts from rest, so its initial speed (\(v_i\)) is 0 m/s.
Mass of the body (\(m\)) = 4.0 kg
\(KE_i = \frac{1}{2} \times m \times v_i^2\)
\(KE_i = \frac{1}{2} \times 4.0 \text{ kg} \times (0 \text{ m/s})^2\)
\(KE_i = \frac{1}{2} \times 4.0 \times 0 \text{ J}\)
\(KE_i = 0 \text{ J}\)
The initial kinetic energy of the body is 0 J.
Final Kinetic Energy
The body gains a final speed (\(v_f\)) of 5 m/s.
Mass of the body (\(m\)) = 4.0 kg
\(KE_f = \frac{1}{2} \times m \times v_f^2\)
\(KE_f = \frac{1}{2} \times 4.0 \text{ kg} \times (5 \text{ m/s})^2\)
\(KE_f = \frac{1}{2} \times 4.0 \times 25 \text{ J}\)
\(KE_f = 2.0 \times 25 \text{ J}\)
\(KE_f = 50 \text{ J}\)
The final kinetic energy of the body is 50 J.
Applying the Work-Energy Theorem
The work done by the constant force is equal to the change in kinetic energy:
\(W = KE_f - KE_i\)
\(W = 50 \text{ J} - 0 \text{ J}\)
\(W = 50 \text{ J}\)
The work done by the force is 50 J.
Understanding Work and Energy Concepts
Work is done when a force causes displacement. In this case, the constant force caused the body to move and increase its speed. The work done by this force is directly related to the energy it transferred to the body, specifically in the form of kinetic energy.
Since the body started from rest (\(KE_i = 0\)), all the work done by the net force (which is the constant force in this scenario) went into increasing its kinetic energy to its final value (\(KE_f\)).
Revision Table: Work and Energy Key Points
| Concept | Formula | Description |
|---|---|---|
| Work Done (W) | \(W = F \cdot d \cdot \cos(\theta)\) (for constant force) or \(W = \Delta KE\) (Work-Energy Theorem) | Energy transferred to an object by a force causing displacement. Unit is Joules (J). |
| Kinetic Energy (KE) | \(KE = \frac{1}{2}mv^2\) | Energy possessed by an object due to its motion. Unit is Joules (J). |
| Work-Energy Theorem | \(W_{net} = \Delta KE = KE_f - KE_i\) | The net work done on an object equals the change in its kinetic energy. |
Additional Information on Constant Force and Work
A constant force means a force that does not change in magnitude or direction over time. When a constant force acts on an object, the work done by that force can be calculated in several ways, depending on the information available.
If the displacement is known and the force is in the direction of displacement (or makes a constant angle), we can use \(W = F \cdot d \cdot \cos(\theta)\). However, if we know the change in speed, using the work-energy theorem (\(W = \Delta KE\)) is often simpler, as shown in this problem.
It's important to note that the work-energy theorem applies to the net work done by all forces. In this specific problem, the problem states "Under the action of a constant force", implying this force is the one causing the change in motion, and potentially the net force if no other forces (like friction, though not mentioned) are doing work.
The work done is a scalar quantity, representing the energy transferred. A positive work done means energy is transferred to the object, increasing its kinetic energy (if it's the net work). A negative work done means energy is transferred away from the object, decreasing its kinetic energy.
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