Physics Q85 – AP-EAMCET 2026

Home Physics Laws of MotionQuestion #208

A block of mass

2 kg

placed on a rough horizontal surface is pulled with a force of

30 N

which makes an angle of

\sin^{- 1} (0.6)

with the horizontal. If the coefficient of kinetic friction between the block and the surface is

0.4

, then the acceleration of the block is (Acceleration due to gravity =

10 {ms}^{- 2}

)

Clue 1 of 3: Resolve Forces

When a force is applied at an angle, it's crucial to resolve it into its horizontal and vertical components to analyze its effect on motion and normal force.

🧠 Full Solution Path

Step 1: Resolve the Applied Force and Determine Vertical Forces✦ Active

The mass of the block is $m = 2 kg$ . The applied force is $F = 30 N$ at an angle $θ = \sin^{- 1} (0.6)$ with the horizontal. This means $\sin θ = 0.6$ . We can find $\cos θ$ using the identity $\sin^{2} θ + \cos^{2} θ = 1$ .

\cos θ = \sqrt{1 - \sin^{2} θ} = \sqrt{1 - (0.06)^{2}} = \sqrt{1 - 0.36} = \sqrt{0.64} = 0.8

The horizontal component of the applied force is $F_{x} = F \cos θ$ , and the vertical component is $F_{y} = F \sin θ$ .

F_{x} = 30 N \times 0.8 = 24 N

F_{y} = 30 N \times 0.6 = 18 N

Now, consider the vertical forces. The weight of the block acts downwards, $W = m g$ . The normal force $N$ acts upwards, and the vertical component of the applied force $F_{y}$ also acts upwards. Since there is no vertical acceleration, the net vertical force is zero.

W = m g = 2 kg \times 10 {ms}^{- 2} = 20 N

N + F_{y} - W = 0

N + 18 N - 20 N = 0

N = 20 N - 18 N = 2 N

💡 Teacher's Secret Hint

Remember to correctly identify the direction of $F_{y}$ . If the force was pushing down, $F_{y}$ would add to the weight, increasing the normal force.

Step 2: Calculate the Kinetic Friction Force○ Expand

The coefficient of kinetic friction is given as $μ_{k} = 0.4$ . Using the calculated normal force $N$ , we can find the kinetic friction force $f_{k}$ .

f_{k} = μ_{k} N

f_{k} = 0.4 \times 2 N = 0.8 N

💡 Teacher's Secret Hint

The friction force always opposes the direction of motion or impending motion. In this case, it opposes the horizontal component of the applied force.

Step 3: Determine the Net Horizontal Force and Acceleration○ Expand

Now consider the horizontal forces. The horizontal component of the applied force $F_{x}$ acts in the direction of motion, and the kinetic friction force $f_{k}$ opposes it. The net horizontal force will cause the acceleration of the block according to Newton's second law ( $F_{n e t, x} = m a$ ).

F_{n e t, x} = F_{x} - f_{k}

m a = F_{x} - f_{k}

2 kg \times a = 24 N - 0.8 N

2 a = 23.2 N

a = \frac{23.2}{2} = 11.6 {ms}^{- 2}

💡 Teacher's Secret Hint

Always ensure your units are consistent throughout the calculation. Here, N (kg·m/s²) is consistent with mass in kg and acceleration in m/s².

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Physics Question 85 – AP-EAMCET 2026

Clue 1 of 3: Resolve Forces

🧠 Full Solution Path