Question

A horizontal force of 10 N is applied on a block of mass 1.5 kg which is initially at rest on a rough horizontal surface. The work done by the applied force in a time of 6 s from the beginning of the motion is (Acceleration due to gravity = 10 ms\(^{-2}\); the coefficient of kinetic friction between the block and the surface is 0.2)

• A. 588 J
• B. 360 J
• C. 840 J
• D. 420 J

Hint:

Be careful to distinguish between the work done by the *applied force* and the *net work* done. Net work would be calculated using the net force (\(W_{net} = F_{net} \times s = 7 \times 84 = 588\) J), which equals the change in kinetic energy. The question specifically asks for the work done by the applied force.

Answer 1

The Correct Option is C

Step 1: Calculate the forces acting on the block.
Applied force, \(F_{app} = 10\) N.
Normal force, \(N = mg = 1.5 \times 10 = 15\) N.
Frictional force, \(F_f = \mu_k N = 0.2 \times 15 = 3\) N. This force opposes the motion.
Step 2: Calculate the net force and the acceleration of the block.
Net force, \(F_{net} = F_{app} - F_f = 10 - 3 = 7\) N.
Acceleration, \(a = \frac{F_{net}}{m} = \frac{7}{1.5} = \frac{14}{3}\) m/s\(^2\).
Step 3: Calculate the distance travelled by the block in 6 seconds.
The block starts from rest (\(u=0\)). We use the kinematic equation \(s = ut + \frac{1}{2}at^2\).
Distance, \(s = 0 \cdot (6) + \frac{1}{2} \left(\frac{14}{3}\right) (6)^2 \).
\(s = \frac{1}{2} \cdot \frac{14}{3} \cdot 36 = 7 \cdot 12 = 84\) m.
Step 4: Calculate the work done by the applied force.
Work done is defined as \(W = F \cdot s\), where F is the force doing the work and s is the displacement in the direction of the force.
Work done by the applied force, \(W_{app} = F_{app} \times s\).
\(W_{app} = 10 \text{ N} \times 84 \text{ m} = 840\) J.