A body of mass 5 kg at rest is acted upon by two forces 30 N and F. If the angle between the forces is 60 degrees and the distance travelled by the body in a time of 2s under the action of these forces is 28m, then F =

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q86 – AP-EAMCET 2026

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Home Physics Laws of MotionQuestion #209

A body of mass

5 kg

at rest is acted upon by two forces

30 N

and

F

. If the angle between the forces is

60^{\circ}

and the distance travelled by the body in a time of

2 s

under the action of these forces is

28 m

, then

F =

Clue 1 of 3: Kinematics and Dynamics Relationship

This problem connects kinematics (motion with constant acceleration) and dynamics (forces and Newton's laws). First, determine the acceleration from the given kinematic information.

🧠 Full Solution Path

Step 1: Calculate the acceleration of the body✦ Active

The body starts from rest ( $u = 0$ ) and travels a distance $s = 28 m$ in time $t = 2 s$ . We can use the kinematic equation for constant acceleration:

s = u t + \frac{1}{2} a t^{2}

Substitute the given values:

28 m = (0) (2 s) + \frac{1}{2} a (2 s)^{2}

Simplifying the equation:

28 = 2 a

Solving for $a$ :

a = \frac{28}{2} = 14 {m/s}^{2}

💡 Teacher's Secret Hint

Ensure units are consistent throughout the calculation. Here, meters and seconds are standard SI units.

Step 2: Calculate the net force acting on the body○ Expand

Using Newton's second law, the net force ( $F_{n e t}$ ) is the product of mass ( $m$ ) and acceleration ( $a$ ). The mass of the body is $m = 5 kg$ .

F_{n e t} = m a

Substitute the values of $m$ and $a$ :

F_{n e t} = (5 kg) (14 {m/s}^{2}) = 70 N

💡 Teacher's Secret Hint

The net force is the vector sum of all forces acting on the object. In this case, it's the resultant of the two applied forces.

Step 3: Formulate the resultant force equation○ Expand

Two forces, $F_{1} = 30 N$ and $F_{2} = F$ , are acting on the body with an angle $θ = 60^{\circ}$ between them. The resultant force $R$ (which is $F_{n e t}$ ) is given by the formula:

R^{2} = F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} \cos θ

Substitute $R = F_{n e t} = 70 N$ , $F_{1} = 30 N$ , $F_{2} = F$ , and $θ = 60^{\circ}$ into the equation:

(70 N)^{2} = (30 N)^{2} + F^{2} + 2 (30 N) (F) \cos (60^{\circ})

Since $\cos (60^{\circ}) = \frac{1}{2}$ :

4900 = 900 + F^{2} + 2 (30) (F) (\frac{1}{2})

Simplifying the equation:

4900 = 900 + F^{2} + 30 F

Rearrange into a quadratic equation:

F^{2} + 30 F + 900 - 4900 = 0

F^{2} + 30 F - 4000 = 0

💡 Teacher's Secret Hint

Remember the trigonometric values for common angles, especially $60^{\circ}$ and $30^{\circ}$ when dealing with force resolution or resultant calculations.

Step 4: Solve for the unknown force F○ Expand

We have a quadratic equation: $F^{2} + 30 F - 4000 = 0$ . We can solve this by factoring or using the quadratic formula. By factoring, we look for two numbers that multiply to -4000 and add to 30. These numbers are 80 and -50.

(F + 80) (F - 50) = 0

This gives two possible solutions for $F$ :

F = - 80 N or F = 50 N

Since the magnitude of a force cannot be negative, we choose the positive value.

F = 50 N

💡 Teacher's Secret Hint

Always check the physical reasonableness of your answers. A negative force magnitude typically indicates an error or a direction opposite to an assumed positive direction, but in this context, magnitude must be positive.

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Physics Question 86 – AP-EAMCET 2026

Clue 1 of 3: Kinematics and Dynamics Relationship

🧠 Full Solution Path