A block of mass $10 ext{ kg}$, moving in the $x$-direction with a constant speed of $10 ext{ m/s}$, is sub

Question

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

Accepted Answer

1. **Calculate Initial Kinetic Energy ($K_i$):**
   Given mass $m = 10 	ext{ kg}$ and initial speed $v = 10 	ext{ m/s}$.
   $$ K_i = \frac{1}{2}mv^2 = \frac{1}{2}(10)(10)^2 = 500 	ext{ J} $$
   (Refer to NCERT Class 11, Section 5.4, Eq. 5.5).
2. **Calculate Work Done by Variable Force ($W$):**
   The force is variable ($F = 0.1x$) and retarding (opposes motion). The work done by a variable force is calculated by integration.
   $$ W = \int_{x_i}^{x_f} F(x) dx $$
   Since it is a retarding force, we take it as negative or subtract the work magnitude from initial energy.
   $$ W = \int_{20}^{30} (-0.1x) dx = -0.1 \left[ \frac{x^2}{2} ight]_{20}^{30} $$
   $$ W = -0.05 (30^2 - 20^2) = -0.05 (900 - 400) $$
   $$ W = -0.05 (500) = -25 	ext{ J} $$
   (Refer to NCERT Class 11, Section 5.5, Eq. 5.7 regarding work done by variable force).
3. **Apply Work-Energy Theorem:**
   The change in kinetic energy is equal to the work done.
   $$ K_f - K_i = W $$
   $$ K_f = K_i + W = 500 + (-25) = 475 	ext{ J} $$
   (Refer to NCERT Class 11, Section 5.6 for the Work-Energy Theorem).

Step-by-Step Solution

Practice All PHYSICS Questions