A boy pushes his friend across a skating rink. Since the frictional forces are very small, the force exerted by the boy on his friend's back is the only significant forece acting on the friend in the horizontal direction. Is the change in kinetic energy of the friend greater than, equal to, or less than the work done by the force exerted by the boy?
It depends upon whether the person who is exerting the force on the friend is also slipping in the opposite direction. If he is on skates and pushing off edges (to use a skating term), he does not slip and the work he performs is converted to kinetic energy of the friend, assuming the friend's friction is negligible.
To determine whether the change in kinetic energy of the friend is greater than, equal to, or less than the work done by the force exerted by the boy, we need to analyze the relationship between work and kinetic energy.
Work is defined as the product of the applied force and the displacement of the object in the direction of the force. Mathematically, it is represented by the equation:
Work = Force × Displacement × cos(θ)
Where θ is the angle between the direction of the applied force and the displacement.
In this case, the force exerted by the boy is the only significant force acting on the friend in the horizontal direction. This means that the displacement of the friend is in the same direction as the force applied by the boy, resulting in θ = 0 degrees. As a result, the cosine of 0 degrees is 1.
Therefore, the equation for work simplifies to:
Work = Force × Displacement
Since the force exerted by the boy on his friend's back is the only significant force acting on the friend in the horizontal direction, the work done by this force is equal to the change in kinetic energy of the friend.
In conclusion, the change in kinetic energy of the friend is EQUAL to the work done by the force exerted by the boy.
To determine whether the change in kinetic energy of the friend is greater than, equal to, or less than the work done by the force exerted by the boy, we need to consider the relationship between work and kinetic energy.
The work done on an object can be calculated using the formula: Work = Force × Distance × cos(θ), where θ is the angle between the force and the direction of the displacement. In this case, since the force is exerted horizontally and the friend is being pushed across the skating rink, the angle θ is 0 degrees, and cos(0) = 1. This simplifies the work formula to: Work = Force × Distance.
The change in kinetic energy of an object can be calculated using the formula: ΔKE = ½ × mass × (final velocity^2 - initial velocity^2). Since the friend is initially at rest (initial velocity = 0), the formula simplifies to: ΔKE = ½ × mass × final velocity^2.
Now, considering the scenario mentioned in the question, the force exerted by the boy on his friend's back is the only significant force in the horizontal direction. This force causes the friend to accelerate and gain speed, resulting in an increased final velocity compared to the initial velocity (0).
Therefore, if the force exerted by the boy does work on the friend, the work done would be positive. In this case, since the force and displacement are in the same direction, the work done by the boy's force is positive.
Since the work done is positive and the friend's final velocity is greater than the initial velocity, the change in kinetic energy of the friend will also be positive. Therefore, the change in kinetic energy of the friend is greater than the work done by the force exerted by the boy.