Question

A horizontal force of 5 N is applied on a stationary body of mass 5 kg, which is initially at rest on a frictionless table. The change in kinetic energy of the body in 10 s is

• A. 25 J
• B. Zero
• C. 125 J
• D. 250 J

Hint:

In problems involving constant force and frictionless surfaces, you can either use kinematic equations to find velocity/displacement and then energy, or use the Work-Energy Theorem directly. Both methods should yield the same result.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem asks for the change in kinetic energy of a body after a constant force acts on it for a specific time, starting from rest on a frictionless surface.
Step 2: Key Formula or Approach:
1. Newton's second law: \( F = ma \Rightarrow a = F/m \).
2. Kinematics (from rest): \( v = u + at \Rightarrow v = at \).
3. Kinetic Energy: \( K = \frac{1}{2}mv^2 \).
4. Work-Energy Theorem: \( \Delta K = W_{net} = F \cdot s \).
Alternatively, we can find the final velocity and then the final kinetic energy.
Step 3: Detailed Explanation:
Given values:
- Force, \( F = 5 \text{ N} \)
- Mass, \( m = 5 \text{ kg} \)
- Initial velocity, \( u = 0 \text{ m/s} \) (at rest)
- Time, \( t = 10 \text{ s} \)
First, calculate the acceleration (\( a \)):
\[ a = \frac{F}{m} = \frac{5 \text{ N}}{5 \text{ kg}} = 1 \text{ m/s}^2 \] Next, calculate the final velocity (\( v \)) after 10 s:
\[ v = u + at = 0 + (1 \text{ m/s}^2)(10 \text{ s}) = 10 \text{ m/s} \] Now, calculate the change in kinetic energy (\( \Delta K \)):
Since it starts from rest, initial kinetic energy \( K_i = 0 \).
Final kinetic energy \( K_f = \frac{1}{2}mv^2 = \frac{1}{2}(5 \text{ kg})(10 \text{ m/s})^2 \)
\[ K_f = \frac{1}{2} \times 5 \times 100 = 250 \text{ J} \] \[ \Delta K = K_f - K_i = 250 \text{ J} - 0 \text{ J} = 250 \text{ J} \]
Step 4: Final Answer:
The change in kinetic energy of the body is 250 J.