Work and Energy - Work done by a constant force

Problem Statement:

A block of mass m = 5 kg moves with friction on a horizontal plane under the action of a constant horizontal force of magnitude F = 40 N. If the block starts from a rest situation and the coefficient of friction with the plan is μ = 0.3, determine its speed when he has traveled a distance d = 8 m.

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Solution:

In this problem we will calculate the work of the different forces acting on a body and link it with the variation of the kinetic energy that this body experiences.

First of all we make a diagram representing the different forces that act on the body:

Where dr is the displacement vector. It is tangent to the trajectory of the body at each point and has the direction of its displacement.

To calculate the velocity of the mass at point B, we will use the equation that relates the variation of kinetic energy with the work of the forces that act on it:

The first member of the above equation is the variation of kinetic energy of the particle when it moves from point A to point B:

The second member is the sum of the work of the forces (each one with its sign) that act on the particle:

Next we will calculate each one of them.

The work of a force is given by:

The line integral is calculated along the trajectory (C) of the particle between the initial (A) and final (B) points. The dot product of the force vector and the displacement vector appears in the integral.

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Work of force F:

First we express the vectors F and dr in term of to their vector components. Both vectors are oriented in the positive direction of the x-axis, so they can be expressed as the product of their magnitude with the unit vector i that provides them a direction:

The substitution in the expression of the work gives:

The dot product of a vector by itself is equal to its squared magnitude. Since i is unitary, the dot product is equal to 1.

On the other hand, we can substitute the bounds of the integral. The integration variable is x, so the lower bound is the x-coordinate of point A and the upper bound is the x-coordinate of point B.


Work of normal force N:

Using the work definition again we have:

The work of the normal force is zero because the dot product of two perpendicular vectors is always zero.


Work of weight P:

Using the work definition again we have:

In this particular case the work of the weight is zero because it is perpendicular to the displacement.


Work of the friction force FR:

Using the work definition again we have:

The work of the friction force is negative because it opposes the motion of the body. Therefore, the body loses kinetic energy (speed).

The friction force magnitude is given by the following expression:

We calculate the normal force magnitude by applying Newton’s second law to the block’s motion.

The projection on the axes represented in the figure gives:

From equation (2) we deduce that in this problem the magnitude of the normal force is equal to the magnitude of the weight. Therefore the work of the friction force is:

We substitute the calculated works in the sum of the works to obtain:

And going back to the expression that relates the variation of kinetic energy with the work we get:

Substituting the givens and after isolating the velocity of the mass at point B we finally obtain:

In the problem we have used g = 10 m/s2

Do not forget to include the units in the results of the problems.

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