# Work on an Inclined Plane: Explanation, Equation, and Problems With Solutions

By a Guy Who Teaches Physics for Fun
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Work on an inclined plane is divided into 2 cases: a) when only conservative force like gravity applies and b) when non-conservative forces like friction or tension apply. The two cases are explained with different equation. Let’s discuss it one by one.

## Work on inclined plane when only gravity force is acting on an object

This case is a simple case in which an object is descending an inclined plane and only affected by the force of gravity without friction or other non-conservative forces. We can calculate the work done on the object by using any of these equations written below.
*You need to realize that the normal force is actually a non-conservative force, but because the normal force and the direction of displacement form a $90^o$ angle, the calculation of work done on the object by normal force would be zero which means it can be ignored.
In this case, the object’s potential energy converts to kinetic energy as the object descends down.
When there is no friction or any other non-conservative force, an object descending an inclined plane experiences the same amount of work as an object in a free fall without friction(same mass and height).
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## Work on an inclined plane when non-conservative force applies

This case is a bit more complicated than the previous case. When friction force or any other non-conservative force applies on the object, we could calculate the work using the following equation.
$W_{n c net}=\Delta E_{p}+\Delta E_{k}$
Description:
\begin{aligned} W_{nc}=&\;\text{net work by all non-consv 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}
You can use this equation for an object which is descending or climbing an inclined plane. You can see an example of using this equation on the slide below.

## Work on inclined plane problems with solutions

Of course, you can not fully understand without studying how we use the equation we have just discussed. Below we present some problems that you can study.

### Problem 1(only conservative force applies)

A block of ice is at the top of an inclined plane that has a height of 5 meters with a slope $30^o$ from horizontal. The block of ice is initially at rest and then slides down the plane. Assume because the block is made of ice, we can neglect the surface friction. Determine the work done on the ice block and determine the final velocity of the ice block.

#### Solution:

Let’s first sketch the forces the ice blocks experience. Please swipe the image above.
Then we calculate the distance traveled by the ice block.
\begin{aligned} s=&\; \frac{\text{tinggi bidang miring}}{\sin 30^{\circ}}\\ s=&\; \frac{5}{1/2}\\ s=&\; 10\;\text{meter} \end{aligned}
We can solve this problem with 3 different work equations. Let’s solve this problem with each of these three equations.
##### Equation 1
\begin{aligned} W\;=&\;mg \sin \theta \;s\\ =&(2)(9.81) \sin 30^o \;(10)\\ =&\;98.1 \;\text{Joule} \end{aligned}
##### Equation 2
\begin{aligned} W\;=&\;-\Delta Ep\\ W\;=&\;-mg(h_2-h_1)\\ =&-(2)(9.81)(0-5)\\ =&\;98.1 \;\text{Joule} \end{aligned}
##### Equation 3
We have obtained the work done on the ice block by using equation 1 and 2. We could use equation 3 to find the final velocity of the ice block.
\begin{aligned} W=&\;\Delta Ek\\ W=&\; \frac{1}{2}m({v_2}^2-{v_1}^2)\\ 98.1=&\;\frac{1}{2}(2)({v_2}^2-0)\\ {v_2}=&\sqrt{98.1}=\;9.9 \;m/s \end{aligned}
Actually, in the case where only gravity force applies and the object is initially at rest, we can also use the equation $v=\sqrt{2gh}$ to find the final velocity of the object. Please try it yourself.

#### Discussion

Thus we have obtained the work done on the ice block. In this problem, we could see all the potential energy the ice block has is transformed into kinetic energy. You could see the energy diagram below.

### Problem 2 (non-conservative force applies)

Now the block is not made of ice but something solid that would make the surface friction significant. The coefficient of kinetic friction of the inclined plane with the block is 0.3. a) Calculate the work done by each force acting on the block. b) Also calculate the change in the kinetic energy of the block. c) Calculate the final velocity of the block.

#### Solution:

In this problem, the non-conservative forces that apply are the frictional force and the normal force.
First, let’s sketch every force that acts on the block. You can slide the image above to see it.
We know from problem 1 that the distance traveled by the block is $s=10\; \text{meter}$.
Then we can calculate the work by each force as follows

#### Work by each force

There are 3 forces that apply, namely gravity (w), friction ($f_k$), and normal(N).
The work done by the normal force is as follows.
\begin{aligned} W_{N}=&N \cos 90 \;s\\ =&N \;(0) \;(10)\\ =&0 \end{aligned}
The work done by the frictional force is as follows.
\begin{aligned} W_{fk}=&f_{k} (\cos 180^o) \;s\\ =&mg(\cos 30^o)\mu_k\;(-1)(10) \\ =&(2)(9.81)(\frac{1}{2}\sqrt{3})(0.3)(-1)(10)\\ =&-50.97 \; \text{Joule} \end{aligned}
The work by the force of gravity is as follows.
\begin{aligned} W_{w}=&mg (\sin 30^o) \;s\\ =&(2)(9.81)(0,5)(10) \\ =&98.1\;\text{Joule} \end{aligned}

#### Change in the kinetic energy of the block

We could find the change in kinetic energy by using the following equation.
\begin{aligned} W_{nc}=&\; \Delta Ep \;+\; \Delta Ek\\ \end{aligned}
We will work the equation to get $\Delta Ek$.
\begin{aligned} W_{fk} + W_N=&\; \Delta Ep\;+\;\Delta Ep\\ -50.97 + 0=&mg(h_2-h_1)+\Delta Ek\\ -50.97 =&(2)(9.81)(0-5)+\Delta Ek\\ -50.97=&\;-98.1+\Delta Ek\\ \Delta Ek=&\;47.13\;\text{Joule} \end{aligned}

#### Block's final velocity

We can find the final velocity of the block by using the equation $\Delta Ek=\frac{1}{2}m({v_2}^2-{v_1}^2)$ and the operation is as follows.
\begin{aligned} \Delta Ek=&\frac{1}{2}m({v_2}^2-{v_1}^2)\\ 47.13=&\frac{1}{2}(2)({v_2}^2-0)\\ v_2=&\sqrt{47.13}\\ =\; &6.87 \;m/s \end{aligned}

#### Discussion

In contrast to problem 1, in this case, not all of the potential energy of the block is converted into kinetic energy. Nearly half of the potential energy of the block(50.97 joules) is released to the environment due to friction. So the kinetic energy of the block when it reaches the end of the inclined plane is only 47.13 Joules. You can see the energy chart of this case below.

### Problem 3(non-conservative forces apply)

A block(50 kg) is pulled by Jason through an inclined plane that has a height of 2.5 meters. The coefficient of kinetic friction of the inclined plane is 0.3. The inclined plane has an angle of inclination $30^o$ The force exerted by Jason’s muscles is 400N. (Assume $g=10m/s^2$)
1. Calculate the work by each force acting on the block
2. Determine the change in kinetic energy experienced by the block.
3. Determine the block’s velocity when it reach the top of the inclined plane.

#### Solution and Dicussion

You can see the solution and discussion of this problem in the article work by non-conservative force.

### Problem 4(non-conservative force applies)

An object climbs an inclined plane at a constant speed from a height of $h_1=2m$ meters to $h_2=6m$. The mass of the object is m=20 Kg. Calculate the work experienced by the object.

#### Solution and discussion

To answer this problem, it is important to define the system and the environment.
If we isolate the object from the gravitational field so that the object’s kinetic energy is the only intrinsic energy the object has then the work that the object experiences is $W=\Delta Ek=0$(remember its velocity is constant).
However, if we don’t isolate the gravitational field from the object so that the potential energy is included as the object’s property, then the work is
\begin{aligned} W=&\Delta Ep\\=&mg(h_2-h_1)\\=&(20)(9.81)( 6-2)\\=&784.8\;\text{Joule} \end{aligned}
This means all the forces that act upon the object increase the potential energy of the object.
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