JEE MAIN

Shortcut Methods and Tricks

Problem 1: A 5 kg block is at rest on a horizontal surface. A 10 N force is applied to the block for 5 s. Find the final velocity of the block.

Shortcut Method:

Use the equation:

v = u + at

where:

• v is the final velocity
• u is the initial velocity (in this case, 0 m/s)
• a is the acceleration
• t is the time

In this problem, we have:

• a = F/m = 10 N / 5 kg = 2 m/s^2
• t = 5 s

Substituting these values into the equation, we get:

v = 0 m/s + 2 m/s^2 * 5 s = 10 m/s

Therefore, the final velocity of the block is 10 m/s.

Problem 2: A 20 kg box is being pushed up a ramp that is inclined at 30 degrees to the horizontal. The coefficient of kinetic friction between the box and the ramp is 0.2. What force is required to move the box up the ramp at a constant speed?

Shortcut Method:

Use the equation:

F = mg sinθ + μmg cosθ

where:

• F is the force required
• m is the mass of the box
• g is the acceleration due to gravity
• θ is the angle of the ramp
• μ is the coefficient of kinetic friction

In this problem, we have:

• m = 20 kg
• g = 9.8 m/s^2
• θ = 30 degrees
• μ = 0.2

Substituting these values into the equation, we get:

F = (20 kg)(9.8 m/s^2) sin30° + (0.2)(20 kg)(9.8 m/s^2) cos30°
= 196 N + 78.4 N = 274.4 N

Therefore, the force required to move the box up the ramp at a constant speed is 274.4 N.

Problem 3: A 100 kg car is traveling at 10 m/s when the brakes are applied. The coefficient of static friction between the tires and the road is 0.7. Find the minimum stopping distance of the car.

Shortcut Method:

Use the equation:

d = v^2 / 2μg

where:

• d is the minimum stopping distance
• v is the initial velocity of the car
• μ is the coefficient of static friction
• g is the acceleration due to gravity

In this problem, we have:

• v = 10 m/s
• μ = 0.7
• g = 9.8 m/s^2

Substituting these values into the equation, we get:

d = (10 m/s)^2 / (2 × 0.7 × 9.8 m/s^2) = 24.5 m

Therefore, the minimum stopping distance of the car is 24.5 m.

Problem 4: A 1 kg ball is thrown vertically into the air with a speed of 20 m/s. Find the maximum height reached by the ball.

Shortcut Method:

Use the equation:

h = v^2 / 2g

where:

• h is the maximum height reached by the ball
• v is the initial velocity of the ball
• g is the acceleration due to gravity

In this problem, we have:

• v = 20 m/s
• g = 9.8 m/s^2

Substituting these values into the equation, we get:

h = (20 m/s)^2 / (2 × 9.8 m/s^2) = 20.4 m

Therefore, the maximum height reached by the ball is 20.4 m.

Problem 5: A 2 kg block is suspended from a spring with a spring constant of 100 N/m. Find the period of oscillation of the block.

Shortcut Method:

Use the equation:

T = 2π √(m / k)

where:

• T is the period of oscillation
• m is the mass of the block
• k is the spring constant

In this problem, we have:

• m = 2 kg
• k = 100 N/m

Substituting these values into the equation, we get:

T = 2π √(2 kg / 100 N/m) = 1.4 s

Therefore, the period of oscillation of the block is 1.4 s.

CBSE BOARD EXAM:

Shortcut Methods and Tricks

Problem 1: A 10 kg block is at rest on a horizontal surface. A force of 20 N is applied to the block for 2 s. Find the final velocity of the block.

Shortcut Method:

Use the equation:

v = u + at

where:

• v is the final velocity
• u is the initial velocity (in this case, 0 m/s)
• a is the acceleration
• t is the time

In this problem, we have:

• a = F/m = 20 N / 10 kg = 2 m/s^2
• t = 2 s

Substituting these values into the equation, we get:

v = 0 m/s + 2 m/s^2 * 2 s = 4 m/s

Therefore, the final velocity of the block is 4 m/s.

Problem 2: A 15 kg box is being pushed up a ramp that is inclined at 30 degrees to the horizontal. The coefficient of kinetic friction between the box and the ramp is 0.3. What force is required to move the box up the ramp at a constant speed?

Shortcut Method:

Use the equation:

F = mg sinθ + μmg cosθ

where:

• F is the force required
• m is the mass of the box
• g is the acceleration due to gravity
• θ is the angle of the ramp
• μ is the coefficient of kinetic friction

In this problem, we have:

• m = 15 kg
• g = 9.8 m/s^2
• θ = 30 degrees
• μ = 0.3

Substituting these values into the equation, we get:

F = (15 kg)(9.8 m/s^2) sin30° + (0.3)(15 kg)(9.8 m/s^2) cos30°
= 147 N + 44.1 N = 191.1 N

Therefore, the force required to move the box up the ramp at a constant speed is 191.1 N.

Problem 3: A 70 kg car is traveling at 15 m/s when the brakes are applied. The coefficient of static friction between the tires and the road is 0.6. Find the minimum stopping distance of the car.

Shortcut Method:

Use the equation:

d = v^2 / 2μg

where:

• d is the minimum stopping distance
• v is the initial velocity of the car
• μ is the coefficient of static friction
• g is the acceleration due to gravity

In this problem, we have:

• v = 15 m/s
• μ = 0.6
• g = 9.8 m/s^2

Substituting these values into the equation, we get:

d = (15 m/s)^2 / (2 × 0.6 × 9.8 m/s^2) = 18.02 m

Therefore, the minimum stopping distance of the car is 18.02 m.

Problem 4: A 0.5 kg ball is thrown vertically into the air with a speed of 15 m/s. Find the maximum height reached by the ball.

Shortcut Method:

Use the equation:

h = v^2 / 2g

where:

• h is the maximum height reached by the ball
• v is the initial velocity of the ball
• g is the acceleration due to gravity

In this problem, we have:

• v = 15 m/s
• g = 9.8 m/s^2

Substituting these values into the equation, we get:

h = (15 m/s)^2 / (2 × 9.8 m/s^2) = 11.9 m

Therefore, the maximum height reached by the ball is 11.9 m.

Problem 5: A 1 kg block is suspended from a spring with a spring constant of 50 N/m. Find the period of oscillation of the block