Physics Q85 – AP-EAMCET 2025

Home Physics Work, Energy & PowerQuestion #643

A body of mass

6 kg

is moved with uniform speed on a rough horizontal surface through a distance of

200 cm

. If the coefficient of kinetic friction between the surface and the body is

0.1

, then the work done against friction is (Acceleration due to gravity =

10 {m s}^{- 2}

)

Clue 1 of 3: Work Done Against Friction

Work done against friction is the energy dissipated due to the opposing force of friction over a certain distance.

🧠 Full Solution Path

Step 1: Identify Given Parameters and Convert Units✦ Active

List the given values and ensure they are in SI units:

m = 6 kg

d = 200 cm = 200 \times 10^{- 2} m = 2 m

μ_{k} = 0.1

g = 10 {m s}^{- 2}

💡 Teacher's Secret Hint

Always convert all given quantities to a consistent system of units (like SI units) at the beginning of the problem to avoid errors in calculation.

Step 2: Calculate the Normal Force○ Expand

For a body on a horizontal surface, the normal force $N$ is equal to the gravitational force (weight) acting on the body.

N = m g

N = 6 kg \times 10 {m s}^{- 2}

N = 60 N

💡 Teacher's Secret Hint

The normal force is perpendicular to the surface. On a flat horizontal surface, it balances the weight of the object if there are no other vertical forces.

Step 3: Calculate the Kinetic Frictional Force○ Expand

The kinetic frictional force $f_{k}$ is the product of the coefficient of kinetic friction $μ_{k}$ and the normal force $N$ .

f_{k} = μ_{k} N

f_{k} = 0.1 \times 60 N

f_{k} = 6 N

💡 Teacher's Secret Hint

Kinetic friction acts when there is relative motion between surfaces. Static friction acts when there is no relative motion.

Step 4: Calculate the Work Done Against Friction○ Expand

The work done against friction $W_{f}$ is the product of the frictional force $f_{k}$ and the distance $d$ over which the force acts.

W_{f} = f_{k} \times d

W_{f} = 6 N \times 2 m

W_{f} = 12 J

💡 Teacher's Secret Hint

Work done against a force is positive if the force applied is in the direction of displacement, effectively overcoming the opposing force. In this case, the work done by the external agent to move the body against friction is $12 J$ .

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Physics Question 85 – AP-EAMCET 2025

Clue 1 of 3: Work Done Against Friction

🧠 Full Solution Path