In Fig.8.51, a block slides down an incline. As it moves from point Ato point B, which are 5.0 m apart, force $\overrightarrow{\mathbf{F}}$ acts on the block, with magnitude 2.0 N and directed down the incline. The magnitude of the frictional force acting on the block is 10 N . If the kinetic energy of the block increases by 35 J between Aand B, how much work is done on the block by the gravitational force as the block moves from Ato B?

The distance between point A and point B, d = 5.0 m

The magnitude of applied force, F = 2.0 N

The magnitude of the friction force, f = 10 N

The increase in kinetic energy,$∆K=35\mathrm{J}$

Use the equation of work relating mechanical energy and thermal energy. Mechanical energy is the sum of kinetic and potential energy. Thermal energy is the energy contained within a system that is responsible for its temperature.

Formulae are as follow:

$$\begin{gathered} W=∆E_{mec}+∆E_{th} \\ ∆E_{mec}=∆K+∆U \\ ∆E_{th}=fd \\ W=F.d \end{gathered}$$

where, $∆K$is change in kinetic energy,$∆U$ is change in potential energy, f, Fare forces,$∆E_{mec},∆E_{th}$ are mechanical and thermal energies, d is displacement and W is work done.

The equation for work done is,

$$\begin{gathered} W=∆E_{mec}+∆E_{th} \\ F.d=∆K+∆U+fd \end{gathered}$$

So,

$$\begin{gathered} F.d=∆K+∆U+fd \\ =2.0\times 5.0-35-10\times 5.0 \\ ∆U=-75\mathrm{J} \end{gathered}$$

The work done by gravitational energy will be equal to the change in gravitational potential energy. So,

$W_g=∆U=-75\mathrm{J}$

Hence,the amount of work done on the block by the gravitational force is$W_g=-75\mathrm{J}$.

Therefore, the work done by gravitation force can be found using the equation of work relating to energy.