Question

A block of mass 0.5 kg is at rest on a horizontal table. The coefficient of kinetic friction between the table and the block is 0.2. If a horizontal force of 5 N is applied on the block, the kinetic energy of the block in a time of 4 s is (Acceleration due to gravity = 10 m/s\(^2\))

• A. \(64 \text{ J}\)
• B. \(128 \text{ J}\)
• C. \(256 \text{ J}\)
• D. \(512 \text{ J}\)

Hint:

Apply Newton's second law to find acceleration and use kinetic energy formula to calculate energy.

Answer 1

The Correct Option is C

Let's solve the problem step-by-step. Given: Mass of the block, $m = 0.5$ kg Coefficient of kinetic friction, $\mu_k = 0.2$ Applied horizontal force, $F = 5$ N Time, $t = 4$ s Acceleration due to gravity, $g = 10$ m/s$^2$ First, calculate the frictional force acting on the block: $F_f = \mu_k \times N = \mu_k \times mg$ $F_f = 0.2 \times 0.5 \times 10 = 1$ N Next, calculate the net force acting on the block: $F_{net} = F - F_f = 5 - 1 = 4$ N Now, calculate the acceleration of the block: $a = \frac{F_{net}}{m} = \frac{4}{0.5} = 8$ m/s$^2$ Calculate the final velocity of the block after 4 seconds: $v = u + at$ Since the block starts from rest, $u = 0$. $v = 0 + 8 \times 4 = 32$ m/s Finally, calculate the kinetic energy of the block: $KE = \frac{1}{2} m v^2 = \frac{1}{2} \times 0.5 \times (32)^2 = 0.25 \times 1024 = 256$ J Therefore, the kinetic energy of the block after 4 seconds is 256 J. Final Answer: The final answer is $\boxed{(3)}$