Question

Two forces are acting on a block of mass \(5\,\text{kg}\) as shown in the figure. If the coefficient of kinetic friction between the block and the surface is \(0.25\), then the net work done in accelerating the block for \(15\) seconds is (Acceleration due to gravity \(=10\,\text{ms}^{-2}\)):

• A. \(4.5\,\text{kJ}\)
• B. \(9\,\text{kJ}\)
• C. \(13.5\,\text{kJ}\)
• D. \(2.25\,\text{kJ}\)

Hint:

For work done over a time interval, often the quickest route is: find acceleration \(\rightarrow\) find final velocity \(\rightarrow\) use \(W=\Delta K\).

Answer 1

The Correct Option is B

Concept: According to the Work-Energy Theorem, \[ W_{\text{net}}=\Delta K \] where \(\Delta K\) is the change in kinetic energy. To calculate the work done, we first determine the acceleration using Newton's second law and then find the final velocity after \(15\) seconds.

Step 1: Determine frictional force. Normal reaction: \[ N=mg=5\times10=50\,N \] Kinetic friction: \[ f_k=\mu N \] \[ f_k=0.25\times50 \] \[ f_k=12.5\,N \]

Step 2: Find the net force from the figure. From the given force configuration, \[ F_{\text{net}}=20\,N \] Hence \[ a=\frac{F_{\text{net}}}{m} \] \[ a=\frac{20}{5} \] \[ a=4\,\text{ms}^{-2} \]

Step 3: Calculate final velocity after 15 s. Assuming the block starts from rest, \[ v=u+at \] \[ v=0+4(15) \] \[ v=60\,\text{ms}^{-1} \]

Step 4: Apply Work-Energy theorem. \[ W=\frac12 mv^2 \] \[ W=\frac12(5)(60)^2 \] \[ W=2.5\times3600 \] \[ W=9000\,J \] \[ W=9\,kJ \]

Step 5: Final answer. \[ \boxed{9\,kJ} \]