Question

A block of mass 10 kg is moving in x-direction with a constant speed of 10 m/s. It is subjected to a retarding force F=0.1xJ/m during its travel from x=20m to x=30cm. Its final kinetic energy will be :

Answer 1

Answer:Answer is given below

Answer 2

Answer:

The final kinetic energy of the block will be 475 J.

Explanation:

Given:   Mass, m = 10 kg

           Speed, v = 10 m/s

           Force, f = 0.1x J/m

         Initial position = 20 m

         Final position = 30 m

To find: The final kinetic energy of the block.

Solution: Since the force is variable, we know that for a variable force:

              $\int\limits dW = \int\limits F. ds$                    

Putting the given values as per the question:

              $W = \lim_{20 \to \ 30} \int\limits 0.1x \, dx$

              $W = \lim_{20 \to \ 30}\int\limits (\frac{1}{10} ) dx^{2}$

              $W = \frac{1}{10} [\frac{x^{2} }{2} ]^{20\, to \, 30}$

              $W = \frac{1}{20} (900-400)$

              $W = \frac{500}{20} = - 25\, J$

Note: Negative is added to the numerical value to show the retarding force in work done.

Now, we know by the relation between work done and kinetic energy that:

                     W = Δ K.E.

So, putting the given values in the formula, we get:

                - 25 = K.E.f - K.E.i

                - 25 = K.E.f - 1/2 × 10 × 10²

                - 25 = K.E.f - 500

              K.E.f = 475 J

Hence, the final kinetic energy of the block will be 475 J.