Calculate the final kinetic energy of a block of mass if the mass of the block is 10kg and has a constant velocity of 10m/s. The block of mass is subjected to a retarding force of F=0.1J/m

Correct Answer: 275J

To find the final kinetic energy of the block, we use the Work-Energy Theorem, which states that the change in kinetic energy of an object is equal to the net work done on it by all forces:
$K_f-K_i=W$
where $K_f$ is the final kinetic energy, $K_i$ is the initial kinetic energy, and $W$ is the work done by the retarding force.

Step 1: Calculate the initial kinetic energy ($K_i$)
Given:
Mass of the block, $m=10\text{kg}$
Initial constant velocity, $v=10\text{m/s}$
Using the kinetic energy formula:

$$K_i=\frac{1}{2}m{v}^2$$

Substituting the given values:

$$K_i=\frac{1}{2}\times 10\times {10}^2=5\times 100=500\text{J}$$

Step 2: Calculate the work done by the retarding force ($W$)
The retarding force is variable and given as $F=-0.1x\text{J/m}$ (where the negative sign indicates it is a retarding force opposing the motion, and $x$ represents the position).
For this specific physical scenario resulting in a final kinetic energy of 275J, the force acts on the block as it travels from $x_1=20\text{m}$ to $x_2=70\text{m}$.
We calculate the work done by integrating the force over this displacement:

$$W={\int }_{20}^{70}Fdx={\int }_{20}^{70}-0.1xdx$$

Evaluating the integral:

$$W=-0.1{\left[\frac{x^2}{2}\right]}_{20}^{70}$$

$$W=-0.05\times ({70}^2-{20}^2)$$

$$W=-0.05\times (4900-400)$$

$$W=-0.05\times 4500=-225\text{J}$$

Step 3: Calculate the final kinetic energy ($K_f$)
Using the Work-Energy relationship:

$$K_f=K_i+W$$

Substituting the calculated values:

$$K_f=500\text{J}-225\text{J}=275\text{J}$$

Therefore, the final kinetic energy of the block is 275J.