Question

You push a \(67-\mathrm{kg}\) box across a floor, where the coefficient of kinetic friction is \(\mu_{\mathrm{k}}=0.55\). The force you exert is horizontal. How much power is needed to push the box at a speed of \(0.50 \mathrm{~m} / \mathrm{s}\) ?

Short answer

Approximately 180.57 watts of power is needed.

Step-by-step solution

Understand the Problem

To calculate the power needed to push the box, we first need to identify the forces acting on the box. The box is moving with a constant velocity across a surface with kinetic friction, which tells us the work done should overcome the frictional force. The power is then the work done per unit time while moving at a constant speed.

Calculate the Normal Force

Since the box is moving horizontally, the normal force is equal to the gravitational force acting on the box. This can be calculated using the formula \( F_{\text{normal}} = m \cdot g \), where \( m = 67 \) kg is the mass of the box and \( g = 9.8 \) m/s² is the acceleration due to gravity. Thus, the normal force is \( 67 \times 9.8 = 656.6 \) N.

Determine the Frictional Force

The frictional force can be calculated using the formula \( F_{\text{friction}} = \mu_{\text{k}} \cdot F_{\text{normal}} \), where \( \mu_{\text{k}} = 0.55 \) is the coefficient of kinetic friction. So, we have \( F_{\text{friction}} = 0.55 \times 656.6 = 361.13 \) N.

Calculate Work Done Against Friction

The work done against friction per unit time, which is the rate at which work is done or power, is given by \( P = F_{\text{friction}} \times v \), where \( v = 0.50 \) m/s is the speed of the box. Substituting the values, the power needed is \( P = 361.13 \times 0.50 = 180.565 \) W.

Key concepts

Power

Power is the rate at which work is done over time. Essentially, it measures how quickly energy is used or transferred from one system to another. It is important as it dictates the efficiency and speed of any activity, from moving a box to powering a factory.
In our scenario, we calculate power by determining how much work is done to move the box at a constant velocity despite opposing friction. The formula to find power, in this case, is given by:

• \( P = F_{\text{friction}} \times v \)
• Where \( P \) is power, \( F_{\text{friction}} \) is the frictional force, and \( v \) is the velocity of the box.

By plugging the frictional force and constant velocity into this equation, we determine the necessary power to maintain movement against friction, highlighting how real-world applications of physics require energy consideration.

Normal Force

Normal force is a reactionary force exerted by a surface to support the weight of an object resting on it. It's always perpendicular to the surface. Understanding this concept is crucial, as it plays a role in calculating frictional forces.
In this exercise, normal force (

• \( F_{\text{normal}} \)

) equals the gravitational force acting on the box, since the box moves horizontally, showing no vertical motion. This is computed as:

• \( F_{\text{normal}} = m \cdot g \)
• Where \( m \) is the mass and \( g \) is the acceleration due to gravity.

The result gives us the force that not only supports the box but also plays a role in determining the friction between the box and the surface.

Frictional Force

Frictional force is the resistance that one surface or object encounters when moving over another. It is crucial in this context as it counters the direction of movement, necessitating an applied force to sustain motion.
This force can be calculated using:

• \( F_{\text{friction}} = \mu_{\text{k}} \times F_{\text{normal}} \)
• Where \( \mu_{\text{k}} \) is the coefficient of kinetic friction.

It highlights how different surfaces interact, where a higher \( \mu_{\text{k}} \) suggests more resistance to motion. Here, understanding this interaction helps calculate the necessary force to counteract friction, enabling smooth and continued movement of our box.

Work Done

Work done refers to the energy transferred when a force moves an object over a distance. It is vital in calculating the necessary energy required to move an object against friction, particularly in ensuring constant movement.
Though our box moves horizontally, the work is done against the frictional force as this energy combats resistance:

• Here, power relates to work done per unit time, thus \( P = F_{\text{friction}} \times v \) aids in finding the energy needed.

Work is a crucial component in physics, shedding light on how forces accomplish movement and how energy conversion happens during the transportation of objects.

Constant Velocity

Constant velocity means that the object is moving at a steady speed in a straight line. Importantly, this implies no net force acts upon it — the forces of friction and applied force are balanced.
This is key to solving our problem. Since the box moves consistently, the applied force equals the frictional force. This constant motion also allows us to use several simplifying assumptions in calculations, focusing solely on overcoming the force of friction:

• No acceleration, which implies energy input directly counters friction.

Understanding constant velocity simplifies many physics problems by indicating a balance of forces and providing a direct application case of Newton's first law.