Net work done by friction = m * g * (μ2 * Distance2 - μ1 * Distance1).
To find the net work done by friction in pulling the board directly from region 1 to region 2, we need to consider the work done by the force of friction in each region separately.
In region 1, the coefficient of kinetic friction is given as μ1. The work done by friction in this region can be calculated using the equation:
Work1 = Force1 * Distance1 * cos(θ1)
Where:
- Force1 is the force of friction in region 1, which is equal to the normal force (N1) multiplied by the coefficient of kinetic friction (μ1). The normal force can be calculated as N1 = m * g, where m is the mass of the board and g is the acceleration due to gravity.
- Distance1 is the distance over which the board moves in region 1.
- θ1 is the angle between the force of friction and the displacement of the board in region 1. Since the board moves directly from region 1 to region 2, θ1 is 0 degrees.
In region 2, the coefficient of kinetic friction is given as μ2. The work done by friction in this region can be calculated using the same equation:
Work2 = Force2 * Distance2 * cos(θ2)
Where:
- Force2 is the force of friction in region 2, which is equal to the normal force (N2) multiplied by the coefficient of kinetic friction (μ2). The normal force can be calculated as N2 = m * g, where m is the mass of the board and g is the acceleration due to gravity.
- Distance2 is the distance over which the board moves in region 2.
- θ2 is the angle between the force of friction and the displacement of the board in region 2. Since the board moves directly from region 1 to region 2, θ2 is also 0 degrees.
To find the net work done by friction, we need to subtract the work done in region 1 from the work done in region 2:
Net work done by friction = Work2 - Work1
Since both θ1 and θ2 are 0 degrees, the cos(θ1) and cos(θ2) terms become 1, simplifying the equation:
Net work done by friction = Force2 * Distance2 - Force1 * Distance1
Substituting the values for the forces of friction in each region:
Net work done by friction = (N2 * μ2 * Distance2) - (N1 * μ1 * Distance1)
Since N1 and N2 are both equal to m * g, the equation becomes:
Net work done by friction = (m * g * μ2 * Distance2) - (m * g * μ1 * Distance1)
Finally, we can express the net work done by friction in terms of m, g, μ1, μ2, Distance1, and Distance2:
Net work done by friction = m * g * (μ2 * Distance2 - μ1 * Distance1)
Complete question :-
A uniform board of length L and mass M, lies near a boundary that separates two regions. In region 1, the coefficient of kinetic friction between the board and the surface is , and in region 2, the coefficient is
. The positive direction is shown in the figure.
Find the net work done by friction in pulling the board directly from region 1 to region 2. Assume that the board moves at a constant velocity,
Express your answer in terms of M, h, L , mu 1 and mu 2.