Question

Jack is holding a box that has a mass of \(m \mathrm{~kg} .\) He walks a distance of \(d \mathrm{~m}\) at a constant speed of \(v \mathrm{~m} / \mathrm{s}\). How much work, in joules, has Jack done on the box? a) \(m g d\) b) \(-m g d\) c) \(\frac{1}{2} m v^{2}\) d) \(-\frac{1}{2} m v^{2}\) e) zero

Short answer

Answer: e) zero

Step-by-step solution

Define Work

(Write the content here) The work done by a force acting on an object can be calculated using the following formula: \(W = F\cdot d\cdot \cos(\theta)\) Where: \(W\) is the work done (in joules), \(F\) is the magnitude of the force applied (in newtons), \(d\) is the distance over which the force is applied (in meters), \(\theta\) is the angle between the direction of motion and the direction of the force (in degrees).

Identify the Force

Jack is holding a box of mass \(m\) kilograms, and hence the force acting on the box due to gravity is \(F_g = mg\). Jack needs to apply a force equal and opposite to the weight of the box so that it does not fall. Thus, the force exerted by Jack on the box in the upward direction is \(F = mg\).

Determine the Angle

While walking, the force exerted by Jack on the box is in the upward direction, while the box moves in the horizontal direction. The angle between the upward force exerted by Jack and the horizontal motion of the box is \(\theta = 90^{\circ}\).

Calculate the Work Done

Now we can plug in the values into the work formula: \(W = F\cdot d\cdot \cos(\theta) = mg\cdot d\cdot \cos (90^{\circ})\) Since \(\cos(90^{\circ}) = 0\), the work done by Jack on the box is: \(W = mg\cdot d\cdot 0 = 0\) So, the correct answer is: e) zero

Key concepts

Work-Energy Principle

The work-energy principle is a fundamental concept in physics that relates the work done on an object to the change in its kinetic energy. Work is defined as the transfer of energy to an object via a force causing the object to move. The formula for work is given by
\( W = F \times d \times \text{cos}(\theta) \),
where \( W \) is the work done, \( F \) represents the magnitude of the force applied, \( d \) is the displacement of the object, and \( \theta \) is the angle between the force and the displacement vectors. When the force is perpendicular to the displacement, no work is done in the direction of the displacement.

In the context of the exercise with Jack, despite exerting effort to hold and carry the box, no work is done on the box in the direction of the motion, because the force he applies is vertical, while the displacement is horizontal. This aligns with the work-energy principle as there is no change in the kinetic energy of the box in the direction of motion, since the speed is constant and the vertical forces cancel out.

Force and Displacement

The concept of force and displacement is crucial to understanding work in physics. Force is a vector quantity, meaning it has both magnitude and direction, while displacement refers to an object's change in position, also a vector. To calculate work, both the magnitude of the force and its direction relative to the displacement matter.

In the given problem, Jack exerts a force equal to the weight of the box (\( mg \)) vertically upwards to counteract the gravitational force pulling downwards. While he moves a certain distance horizontally, the displacement of the box is in the same horizontal direction. Jack's force does not contribute to the displacement since it's at a \(90^\circ\) angle to the movement; hence, even with force applied over a distance, the resulting work done is zero because the force does not aid in the box's horizontal motion.

Angle of Application

The angle of application is the angle between the direction of the applied force and the direction of displacement. The work done by the force is maximum when this angle is zero because the force is in the same direction as the displacement. As the angle increases, the component of force 'useful' for doing work decreases, reaching zero when the angle is \(90^\circ\).

To illustrate, if Jack were pushing the box forward (the same direction as the displacement) instead of carrying it, the angle of application would be \(0^\circ\), and all of his pushing force would contribute to the work done on the box. However, in our scenario, the upward force that Jack applies does not contribute to the displacement of the box horizontally; the angle of application (\( \theta = 90^\circ \)) results in a cosine factor of zero. Consequently, no work is done, as indicated by the exercise solution, signaling the importance of both magnitude and direction of force in work calculations.