How Much Energy Is Needed to Drag a Block Over a Distance?

[indie452]

Homework Statement

whats the energy required to drag a block a distance L

Homework Equations

coefficiant of friction between block and floor = $$\mu$$
density of block = p

force*distance = change in kinetic energy

The Attempt at a Solution

friction acting against movement Fr = $$\mu$$mg m = mass of block

force in direction of movement F = ma

resultant force = F - Fr = m(a-$$\mu$$g)

resultant force*L = energy = m(a-$$\mu$$g)L

i don't think the last step is right though as I am not sure sure that the assumption made that energy required is the same as the change in KE is correct...

help or guidance appreciated

 

[PhanthomJay]

Although not stated as such, I think you are looking for the energy (work) required by a person or machine to just overcome friction and move the block at constant speed (no acceleration).

 

[nlightner]

Your attempt at a solution is on the right track. The energy required to drag a block a distance L is equal to the work done by the external force in overcoming the friction force. The work-energy theorem states that the work done by a force is equal to the change in kinetic energy of the object. In this case, the external force is the force applied to the block to move it, and the change in kinetic energy is the energy required to drag the block a distance L.

So, we can write:

Work done by external force = Change in kinetic energy

F*L = 1/2*m*v^2

where F is the external force, L is the distance dragged, m is the mass of the block, and v is the final velocity of the block after being dragged a distance L.

Now, we can substitute for the external force and the final velocity:

ma*L = 1/2*m*v^2

We can rearrange this to solve for the energy required:

Energy required = m*(v^2/2a)*L

To find the final velocity, we can use the kinematic equation:

v^2 = u^2 + 2as

where u is the initial velocity, which we can assume to be zero since we are starting from rest. So, we have:

v^2 = 2as

Substituting this into our expression for energy required, we get:

Energy required = m*(2as/2a)*L

= m*s*L

= m*g*\mu*L

= m*g*\mu*L

where s is the distance dragged, L is the length of the block, m is the mass of the block, g is the acceleration due to gravity, and \mu is the coefficient of friction between the block and the surface it is being dragged on.

So, the energy required to drag a block a distance L is equal to the mass of the block multiplied by the product of the acceleration due to gravity and the coefficient of friction, all multiplied by the distance dragged.

I hope this helps! Let me know if you have any further questions.