# Work by Non-conservative Force: Explanation, Equation, Problems and Solutions

By a Guy Who Teaches Physics for Fun Choose language:

## What is a Non-Conservative Force

A non-conservative force is a force that when applied to an object, the work experienced by the object depends on the path it takes.
Examples of non-conservative forces are friction force, rope tension, the force exerted by hand when you push an object, normal force, and many more.
Unlike conservative force, non-conservative force does not induce potential energy. You could see the table below. ## Work by non-conservative force

The work an object experienced due to non-conservative forces depends on the path the object took. Let’s dive deeper by using an example. For example, Andi and Wahyu push boxes with different paths as shown in the following picture. Try to answer the following questions. What are the forces that apply to the box? Who put the most work on the box?
There are 4 forces that apply to the box, namely pushing force by hands, weight, normal force, and friction force. Three of these forces, namely the pushing force, normal force, and frictional forces are non-conservative forces.
Normal force and weight create 90 degrees angle with the direction the 2 boxes move. So the work done by both forces is zero. The rest 2 forces, namely the pushing force and the friction force(both non-conservative), are the only 2 forces that do work on the 2 boxes.
The pushing force by hand will do work on the box and that is $W_{hand}=F_{hand}\;s$. While the friction force would also do work on the box and that is $W_{friction}=f_{k}\;s$. You can see from the equations, work by friction force and hand force depend on the value of s(distance). If the path is long in distance then the work done on the box becomes greater.
So, if we look back at the image. It is clear that Wahyu done more work on the object because the path he choose is longer in distance.
The equation we use to calculate work by non-conservative force is as follows.
$W_{nc}=F_{nc}s\cos{\theta}$
Description: $W_{nc}=$ Work by non-conservative force
When more than one non-conservative force applies, the equation becomes the following.
$W_{ncnet}=W_{nc1}+W_{nc2}+W_{nc3}+……..$
Description: $W_{ncnet}=$ The net work by all non-conservative forces that apply on the object
You must be careful in determining when to input the work as positive, negative, or zero while using the equation above. ## Work on an inclined plane when both Conservative and non-conservative forces apply simultaneously

An object can experience work due to conservative and non-conservative forces simultaneously. In this case, we could use the following equation.
$W_{ncnet}=\Delta E_{p}+\Delta E_{k}$
Description: \begin{aligned} W_{ncnet}=&\text{the net work by all nonconsv forces} \\ \Delta E_{p}=&mg{(h_2-h_1)}\\ \Delta E_{k}=&\frac{1}{2}m({v_2}^2-{v_1}^2)=\sum F_x \;s \end{aligned}
The examples below are cases where we could use the equation above:
• An object slides down an inclined plane with friction applies.
• Free fall with air resistance
• A vertically upward motion by an object attached to a pulley

## Problems with solutions to learn

You need to look at some problems below to comprehend the concept of work by when non-conservative force applies better. Calculating the work done on an object when non-conservative force applies is a bit complicated, you may need to read this section slowly.

### Problem 1

A block(2 kg) is at the top of an inclined plane which is 5 meters high. The plane has a $30^{\circ}$ slope angle. The block is initially stationary and then slides down. The coefficient of kinetic friction is known to be 0.3. Determine the speed of the block when it reach the bottom.

#### Solution:

Let’s first sketch all the forces that acted on the block. Please swipe the image above.
Next, we identify the conservative and non-conservative forces acting on the block.
• Conservative force : weight (mg).
• Non-conservative force : Friction $(f_{k})$ and normal force (N).
Then we determine the distance traveled by the block.
\begin{aligned} s=&\; \frac{\text{height}}{\sin 30^{\circ}}\\ s=&\; \frac{5}{1/2}\\ s=&\; 10\;\text{meter} \end{aligned}
Next, we use the work equation for an object which experiences conservative and non-conservative forces simultaneously.
\begin{aligned} W_{nc}=&\; \Delta Ep \;+\; \Delta Ek\\ \end{aligned}
We operate the equation to get the final velocity of the block.
Let’s calculate W_{nc}.
\begin{aligned} W_{nc}=&f_{k} \;s\;\cos 180^o \; + \;N\;s\;\cos 90^o\\ =&N\;\mu_k \;s\;(-1) \; + 0\\ =&(2)(9.81) (\cos 30^o)\;(0.3) \;(10)\;(-1) \; \\ =&50.97\;\text{Joule} \end{aligned}
Let’s find $\Delta Ep$ next.
\begin{aligned} \Delta Ep=&mg(h_{2}-h_{1})\\ =&(2)(9.81)(0-5)\\ =&-98.1\;\text{Joule} \end{aligned}
Let’s also find $\Delta Ek$ next.
\begin{aligned} \Delta Ek=&\frac{1}{2}m({v_{t}}^2-{v_{o}}^2)\\ =&\frac{1}{2}(2)({v_{t}}^2-0)\\ =&\;{v_{t}}^2 \end{aligned}
Now let’s combine all the information we have to find $v_t$.
\begin{aligned} W_{nc}=&\; \Delta Ep \;+\; \Delta Ek\\ -50.97=&\;-98.1+{v_{t}}^2\\ v_{t}=&\;\sqrt{47.13}\\ v_{t}=&\;6.87 \;\text{m/s} \end{aligned}
\begin{aligned} f_{k} \;s\;\cos 180^o \; + \;N\;s\;\cos 90^o=&\; mg(h_{2}-h_{1})\;+\;\frac{1}{2}m({v_{t}}^2-{v_{o}}^2)\\ N\;\mu_k \;s\;(-1) \; + 0=&\; mg(h_{2}-h_{1})\;+\;\frac{1}{2}m({v_{t}}^2)\\ (2)(9.81) (\cos 30^o)\;(0.3) \;(10)\;(-1) \; =&\; (2)(9.81)(0-5)\;+\;\frac{1}{2}(2)({v_{t}}^2)\\ -50.97=&\;-98.1+{v_{t}}^2\\ v_{t}=&\;\sqrt{47.13}\\ v_{t}=&\;6.87 \;\text{m/s}\\ \end{aligned}

#### Discussion

Thus we have obtained the final velocity of the block. We can also represent the work and energy of the block with a bar chart for better understanding.
Slide the chart above and you will see another chart. The second chart represents the work and energy if the block slides without friction. It implies all the potential energy of the block will turn into kinetic energy. So the block would have a higher speed when it reached the bottom of inclined plane.
The first chart represents the work and energy when friction is applied to the block. You can see from the chart, not all of the block’s potential energy is converted into kinetic energy. Close to half of it turns into thermal energy due to friction. In number, 50.97 Joule of energy got absorbed by the surface of the plane and the block. And the rest 47.13 Joule of energy becomes the kinetic energy of the block. That’s why the block would have a lower speed when it reached the bottom.

### Problem 2 Look at the image above. Jason pulls a 50 kg block up an inclined plane. The object travels a distance of 5 meters and moves from bottom to 2.5 meters high. The inclined plane has an angle of $30^o$ and its coefficient of kinetic friction is 0.3. The force exerted by Jason’s muscles is 400N. ($g=10m/s^2$)
1. Identify the conservative and non-conservative forces acting on the block.
2. Calculate the work done by each force acting on the object
3. Determine the change in kinetic energy of the block.
4. Determine the block’s speed at the top of the inclined plane.

#### Solution:

##### 1. Identify conservative and non-conservative forces
• Conservative force: Weight of the block .
• Non-conservative forces: Jason’s pulling force, friction force, and normal force.
##### 2. Calculate the work by each force
\begin{aligned} W_{Jason}=& F.s\\=&(400)(5)\\=&2000\text{Joule} \end{aligned}
\begin{aligned} W_{friction}=&\;N.\mu.s\\=&\;mg\cos \theta \;(0.3)(5)\\ =&\;(50)(10)(\cos 30^o)(0.3)(5)\\ =&\;649.5 \;\text{Joule} \end{aligned}
\begin{aligned} W_{berat}=&\;(mg)(\cos 90^o+\theta)(s)\\ =&\;(50.10)(\cos 90^o+30^o)(5)\\ =&\;(500)(-0.5)(5)\\ =&\;-1250 \;\text{Joule} \end{aligned}
\begin{aligned} W_{Normal}=&(N\cos 90).s\\ =&\;0 \end{aligned}
##### 3. Determine the change in kinetic energy
\begin{aligned} W_{nc}=&\; \Delta Ep \;+\; \Delta Ek\\ W_{Jason}+W_{friction}=&\;mg(h_2-h_1)+\Delta Ek\\ 2000+(-649.5)=&\;(50)(10)(2.5-0) \;+\Delta Ek\\ \Delta Ek=&\;2000-649.5-1250\\ \Delta Ek=&\;100.5 \;\text{Joule} \end{aligned}
##### 4. Determine the block's velocity
\begin{aligned} \Delta Ek =&\;\frac{1}{2}m({v_2}^2-{v_1}^2)\\ 100.5=&\;\frac{1}{2}(50)({v_2}^2-0)\\ v_2=&\sqrt{\frac{100.5}{25}}\\ v_2\approx & \;2\;\text{m/s} \end{aligned}

#### Discussion

We have just found the solution for problem 2. We can still comprehend this problem in more depth by making a work and energy chart as follows. Let’s look at the chart above. Jason spends 2000 Joules of his energy to give work on the block. A total of 1250 Joules of that energy is used to increase the potential energy of the block. A total of 649.5 Joules are used to resist the frictional force and it turned into heat energy. And the rest 100.5 Joule of energy turned into kinetic energy of the block. So, my dude Jason is so strong not only he could pull a 50 kg up the the inclined plane he also gives the block some speed.
Share Button
Related post    