Question

A body of mass 2 kg at rest starts to move under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction, 0.1. The kinetic energy of the body in 10 seconds is (\(g = 10 \text{ ms}^{-2}\))

• A. 1225 J
• B. 100 J
• C. 2025 J
• D. 625 J
• E. 725 J

Hint:

When mass is 2 kg, kinetic energy numerically equals the square of velocity (\(KE = v^2\)) since the \(1/2\) and \(m\) cancel out. This saves calculation steps.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The net force acting on the body is the difference between the applied horizontal force and the frictional force. This net force produces acceleration, which changes the velocity over time.

Step 2: Key Formula or Approach:
1. Frictional force \(f_k = \mu_k N = \mu_k mg\).
2. Net Force \(F_{net} = F_{applied} - f_k\).
3. Acceleration \(a = F_{net} / m\).
4. Final velocity \(v = u + at\).
5. Kinetic Energy \(KE = \frac{1}{2}mv^2\).

Step 3: Detailed Explanation:

Step 1: Calculate the kinetic frictional force.
\[ f_k = 0.1 \times 2 \times 10 = 2 \text{ N} \]

Step 2: Find the net force and resulting acceleration.
\[ F_{net} = 7 \text{ N} - 2 \text{ N} = 5 \text{ N} \]
\[ a = \frac{5}{2} = 2.5 \text{ ms}^{-2} \]

Step 3: Calculate the velocity after 10 seconds (starting from rest, \(u = 0\)).
\[ v = 0 + 2.5 \times 10 = 25 \text{ ms}^{-1} \]

Step 4: Calculate the kinetic energy.
\[ KE = \frac{1}{2} \times 2 \times (25)^2 \]
\[ KE = 625 \text{ Joules} \]

Step 4: Final Answer:
The kinetic energy after 10 seconds is 625 J.