Question

A force of \( F = 0.5 \) N is applied on the lower block as shown in the figure. The work done by the lower block on the upper block for a displacement of 3 m of the upper block with respect to the ground is (Take, \( g = 10 \) m/s\( ^2 \)): 
 

• A. \( -0.5 \) J
• B. \( 0.5 \) J
• C. \( 2 \) J
• D. \( -2 \) J

Hint:

If two blocks are moving together, the acceleration of the system is the same. Work done by friction can be calculated as \( W = f d \).

Answer 1

The Correct Option is B

Step 1: {Calculate maximum acceleration} 
The maximum acceleration that the \( 1 \) kg block can have is: \[ a_{\max} = \mu g = (0.1)(10) = 1 { m/s}^2 \] 
Step 2: {Calculate common acceleration} 
The acceleration of the system is: \[ a = \frac{F}{m_{{total}}} = \frac{0.5}{3} = \frac{0.5}{3} { m/s}^2 \] Since \( a<a_{\max} \), the blocks move together. 
Step 3: {Find force of friction} 
The friction force acting on the upper block: \[ f = m a = (1) \times \frac{0.5}{3} = \frac{1}{6} { N} \] 
Step 4: {Find work done by friction} 
\[ W = f \times d = \frac{1}{6} \times 3 = 0.5 { J} \] Thus, the correct answer is (B) 0.5 J.