Ask Question
6 January, 12:47

A child pushes horizontally on a box of mass m which moves with constant speed v across a horizontal floor. The coefficient of friction between the box and the floor is μ. At what rate does the child do work on the box?

+3
Answers (2)
  1. 6 January, 12:51
    0
    Rate of child doing work on box = μmgv (Unit is Watt)

    Explanation:

    Rate of child doing work on box = Work done / time = Power

    Power = Horizontal force x Velocity

    We are aware that the Velocity in this case is v.

    As the object is moving with constant velocity, the acceleration would be zero and the applied horizontal force will be equal to friction force. So in our case,

    Horizontal force = friction force

    We know that the coefficient of friction is the ratio of friction force to Normal force,

    μ = friction force / Normal force

    Normal Force = mg, where m is the mass and g is the gravitational acceleration

    Friction force = μ x Normal Force

    Friction force = μmg

    Power = μmgv (Unit of power is Watt)
  2. 6 January, 13:00
    0
    P = μ*mg*v

    Explanation:

    A child pushes horizontally on a box of mass m which moves with constant speed v across a horizontal floor. The coefficient of friction between the box and the floor is μ. The rate at which the child works is calculated as shown below:

    mass of the box = m; coefficient of friction is μ; speed = v.

    In order to push the box, the child must exert a force equal to or more than the frictional force.

    force = coefficient of friction*weight of the box

    f = μ*mg

    In addition, to calculate the rate of work (i. e. power). We have:

    Power = force*velocity (or speed)

    Therefore:

    P = μ*mg*v
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “A child pushes horizontally on a box of mass m which moves with constant speed v across a horizontal floor. The coefficient of friction ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers