Question

A block enters a rough horizontal surface with a speed of \( 6 \, \text{ms}^{-1} \) at \( x = 1.5 \, \text{m} \) and leaves the rough horizontal surface with a speed of \( 4 \, \text{ms}^{-1} \) at \( x = 2.5 \, \text{m} \). If the retarding force acting on the block is \( F = -25x \, \text{N} \) (where \( F \) is in newton and \( x \) is in meter), then the mass of the block is:

• A. 2.5 kg
• B. 10 kg
• C. 5 kg
• D. 4 kg

Hint:

The work-energy principle links the net work done by forces on a body to its change in kinetic energy, allowing the determination of mass when forces and displacements are known.

Answer 1

The Correct Option is C

- Calculate the initial and final kinetic energies: \[ KE_1 = \frac{1}{2} m (6)^2 = 18m \, \text{J}, \quad KE_2 = \frac{1}{2} m (4)^2 = 8m \, \text{J} \]
- Work done by the retarding force: \[ W = \int_{1.5}^{2.5} -25x \, dx = -\frac{25}{2} [(2.5)^2 - (1.5)^2] = -50 \, \text{J} \]
- Change in kinetic energy \( \Delta KE = KE_2 - KE_1 = 8m - 18m = -10m \, \text{J} \)
- Equate the work done to the change in kinetic energy: \[ -10m = -50 \Rightarrow m = 5 \, \text{kg} \]