Question

A block of mass 10 kg is moving in x-direction with a constant speed of $10 \mathrm {~ms} ^ { - 1 }$. It is subjected to a retarding force during its travel from x = 20 m to x = 30 m. Its final kinetic energy will be:

• A. 250 J
• B. 275 J
• C. 450 J
• D. 475 J

Answer 1

Answer: (D)

475 J

Answer 2

Here, $\mathrm { m } = 10 \mathrm {~kg} , \mathrm { v } _ { \mathrm { i } } = 10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$

Initial kinetic energy of the block is

$\mathrm { k } _ { \mathrm { i } } = \frac { 1 } { 2 } \mathrm { mv } _ { \mathrm { i } } ^ { 2 } = \frac { 1 } { 2 } \times ( 10 \mathrm {~kg} ) \times \left( 10 \mathrm {~ms} ^ { - 1 } \right) ^ { 2 } = 500 \mathrm {~J}$

Work done by retarding force

$= - 0.1 \left[ \frac { 900 - 400 } { 2 } \right] = - 25 \mathrm {~J}$

According to work-energy theorem