You observe someone pulling a block of mass 43kg across a low-friction surface. While they pull a distance of 3m in the direction of motion, the speed of the block changes from 5 m/s to 7 m/s. Calculate the magnitude of the force exerted by the person on the block. What was the change in internal energy (chemical energy plus thermal energy) of the person pulling the block?

Energy is a term that refers to a person's ability to do tasks. Potential energy, kinetic energy, thermal energy, electrical energy, chemical energy, nuclear energy, and other kinds of energy are all examples of potential energy. There is also heat and work (energy transmitted from one body to another).

Work is the energy delivered to or from an item by applying force along a displacement in physics. It is frequently represented as the product of force and displacement in its simplest form.

The system has only kinetic energy.

The work done W on the block equals the system's change in energy. Write the equation.

$$\begin{gathered} W=E_{f-}{E}_j \\ =\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_j^2 \\ =\frac{1}{2}m{v}_f^2-v_j^2 \end{gathered}$$

Where, m is the mass, $v_f^2-v_j^2$is the change in velocity.

Substitute $m=43kg$, $v_i=7m/s$ and $v_i=5m/s$ into the obtained formula.

$$\begin{gathered} w=\frac{1}{2}\left(43kg\right)\left[{\left(7m/s\right)}^2-{\left(5m/s\right)}^2\right] \\ 516J \end{gathered}$$

Therefore, the work done is $516J$.

Multiply force Fand the displacement $\nabla x$ to find the work.

$W=F\nabla x$

Substitute $W=516J$and $\nabla x=3$ into the obtained formula and solve for F.

$$\begin{gathered} F=\frac{W}{\nabla x} \\ =\frac{516J}{3M} \\ =172N \end{gathered}$$

Therefore, the force is 172N .

The person's final internal energy is equal to their initial energy minus the effort they did on the block. Write the equation.

$$\begin{gathered} E_{internal,f}=E_{{}_{internal,i}}-W \\ \nabla E_{internal}=-W \\ =-516J \end{gathered}$$

Therefore, the force final Internal energy is $-516J$.