JEE Main

1. A 5 kg block is at rest on a horizontal surface. A force of 10 N is applied to the block for 2 s. What is the velocity of the block after 2 s? Shortcut Method:

• Formula: $$V=u+at$$ where,

• V = Final velocity

• U = Initial velocity

• a = acceleration

• t = Time

• Given:

• Initial velocity (u) = 0 m/s

• Force applied (F) = 10 N

• Mass of the block (m) = 5 kg

• Time (t) = 2 s

• Calculation:

• First, we need to find the acceleration (a) using Newton’s second law: F = ma $$a = F/m$$ $$a = 10 N / 5 kg$$ $$a = 2 m/s²$$

Now we can put back the values in the velocity formula: $$V=u+at$$ $$V = 0 m/s + 2 m/s² × 2 s$$ $$V = 4 m/s$$

Therefore, the velocity of the block after 2 seconds is 4 m/s.

2. A 20 kg object is moving at a velocity of 10 m/s. What is the force required to stop the object in 5 s? Shortcut Method:

• Formula: $$F = ma$$ where, F = Force required m = mass of the object a = retardation (negative acceleration)

• Calculation:

• In order to stop the object, we need to apply a force opposite to its direction of motion. So, the acceleration (a) will be negative: $$a = -10 m/s / 5 s$$ $$a = -2 m/s²$$

Now, we can put these values in Newton’s second law to find the force (F) required: $$F = ma$$ $$F = 20 kg × (-2 m/s²)$$ $$F = -40 N$$

Therefore, the force required to stop the object in 5 seconds is -40 N.

Note: The negative sign indicates that the force should be applied in the opposite direction to the object’s motion.

3. A 1000 kg car is moving at a velocity of 20 m/s. What is the force required to stop the car in 10 s? Shortcut Method:

• Formula: $$F = ma$$ where, F = Force required m = mass of the car a = retardation (negative acceleration)

• Calculation:

• Similar to the previous question, we need to find the acceleration (a) first: $$a = -20 m/s / 10 s$$ $$a = -2 m/s²$$

Now, we can put these values in Newton’s second law to find the force (F) required: $$F = ma$$ $$F = 1000 kg × (-2 m/s²)$$ $$F = -2000 N$$

Therefore, the force required to stop the car in 10 seconds is -2000 N.

Note: Again, the negative sign indicates that the force should be applied opposite to the object’s motion.

CBSE Board Exam

1. State and explain Newton’s First Law of motion with examples.

Newton’s First Law of Motion (Law of Inertia):

• Every object remains in a state of rest or of uniform motion in a straight line unless it is acted upon by an external force.

In simpler terms, an object at rest stays at rest, and an object in motion stays in motion with an unvarying velocity, unless an external force is applied.

Examples of Newton’s First Law:

• A book resting on a table remains at rest until someone picks it up or applies a force to move it.
• A car moving on a straight road at a constant speed will continue to do so until the driver applies the brakes or turns the steering wheel, which are external forces acting on the car.
• A ball thrown into the air will keep moving in a straight line (ignoring air resistance) until gravity pulls it back to the ground.

2. A 10 N force is applied to a 2 kg mass. Calculate the acceleration produced.

Shortcut Method:

• Formula: $$a = F/m$$ where, a = acceleration produced F = force applied m = mass of the object

• Calculation: $$a = 10 N / 2 kg$$ $$a = 5 m/s²$$

Therefore, the acceleration produced when a 10 N force is applied to a 2 kg mass is 5 m/s².