JEE Main

1. A block of mass 10 kg is resting on a horizontal surface. A horizontal force of 20 N is applied to the block. The coefficient of static friction between the block and the surface is 0.5. What is the maximum value of the force of friction that can act on the block?

Shortcut Method:

The maximum force of friction that can act on the block is given by:

$$F_{max} = μ_s mg$$

Where,

• (F_{max}) is the maximum force of friction
• (\mu_s) is the coefficient of static friction
• (m) is the mass of the block
• (g) is the acceleration due to gravity

Substituting the given values, we get:

$$F_{max} = 0.5 * 10 * 9.8 = 49 N$$

Therefore, the maximum value of the force of friction that can act on the block is 49 N.

2. A block of mass 10 kg is resting on a horizontal surface. A horizontal force of 20 N is applied to the block. The coefficient of kinetic friction between the block and the surface is 0.4. What is the acceleration of the block?

Shortcut Method:

The acceleration of the block is given by:

$$a = \frac{F - f}{m}$$

Where,

• (a) is the acceleration of the block
• (F) is the applied force
• (f) is the force of friction
• (m) is the mass of the block

Substituting the given values, we get:

$$a = \frac{20 - (0.4 * 10 * 9.8)}{10} = 0.84 m/s^2$$

Therefore, the acceleration of the block is 0.84 m/s2.

3. Two blocks of masses 10 kg and 20 kg are connected by a massless string. The 10 kg block is resting on a horizontal surface. The coefficient of static friction between the 10 kg block and the surface is 0.5. What is the maximum force that can be applied to the 20 kg block before the 10 kg block starts to move?

Shortcut Method:

The maximum force that can be applied to the 20 kg block before the 10 kg block starts to move is given by:

$$F_{max} = μ_s mg’$$

Where,

• (F_{max}) is the maximum force, -(\mu_s) is the coefficient of static friction between the 10 kg block and the surface, -(m) is the mass of the 10 kg block, -(g’) is the effective acceleration due to gravity
  $$g’= g - \frac{F_{20}}{m_{20}}$$ (F_{20}) is the force applied to the 20 kg block $$T=F_{20}$$ substituting this tension T in the above equation of (g’), we get $$g’=g- \frac{F_{20}}{m_{20}}$$ $$g’ =g- \frac{F_{20}}{20}$$

Substituting the value we get $$F_{max}=0.5(10)(g-\frac{F_{20}}{20})$$ This is the condition of impending motion or limiting equilibrium of the 10 kg mass

For equilibrium of the 20 kg mass we know $$\Sigma F=ma$$ (T=ma) as 20 kg mass is free to move $$F_{20}=m_{20}a$$ Substituting the value of (F_{20}) we get $$F_{max}=50(g-\frac{2a}{g})$$ $$F_{max}=50g-10a$$ Differentiate both sides with respect to (a) we get $$\frac{dF_{max}}{da}=-10$$ For maximum value, put (\frac{dF_{max}}{da}=0) we get $$a=5 m/s^2$$ Substituting this value in (F_{max}), we get $$F_{max}=250 N$$ Therefore, the maximum force that can be applied to the 20 kg block is 250 N.

4. Two blocks of masses 10 kg and 20 kg are connected by a massless string. The 10 kg block is resting on a horizontal surface. The coefficient of kinetic friction between the 10 kg block and the surface is 0.4. What is the acceleration of the blocks when a force of 30 N is applied to the 20 kg block?

Shortcut Method:

Let the acceleration of the system of blocks be (a).

For the 10 kg block, the equation of motion is:

$$F - f_{k1} - T = m_1a$$ $$F- (\mu_k m_1g+T)=m_1a \tag1$$

For the 20 kg block, the equation of motion is:

$$T - m_2g = m_2a$$

$$T = m_2(g+a) \tag2$$ Substituting equation (2) into equation (1), we get:

$$F - \mu_k m_1g -m_2(g+a)=m_1a$$

$$F = \mu_k m_1g + m_2g + (m_1 + m_2)a$$

Substituting numerical values $$a=\frac{30-0.4(10)(9.8)-20(9.8+a)}{10+20}$$ $$\frac{1}{30} a=\frac{12.4-2a}{30}$$ $$a=\frac{3}{8}=0.375m/s^2$$ Therefore, the acceleration of the blocks is (0.375m/s^2).

CBSE Board

1. A block of mass 10 kg is resting on a horizontal surface. A horizontal force of 20 N is applied to the block. The block moves a distance of 10 m. What is the work done by the force?

Shortcut Method:

The work done by the force is given by:

$$W = Fd$$ Where,

• (W) is the work done,
• (F) is the applied force,
• (d) is the displacement.

Substituting the given values, we get:

$$W = 20 * 10 = 200 J$$ Therefore, the work done by the force is 200 J.

2. A block of mass 10 kg is moving at a velocity of 5 m/s. The block is brought to rest by a force of 20 N. What is the work done by the force?

Shortcut Method:

The work done by the force is given by:

$$W = F \Delta x $$

Where,

• (W) is the work done,
• (F) is the applied force,
• (\Delta x) is the displacement of the block.

We can find the displacement by using the equation of motion:

$$v^2 = u^2 + 2as $$ Where,

• (v) is the final velocity of the block,
• (u) is the initial velocity of the block,
• (a) is the acceleration of the block,
• (s) is the displacement.

Substituting the given values, we get:

$$0 = 5^2 + 2 * (-20) * (\Delta x) $$

Solving for (\Delta x), we get: $$\Delta x= \frac{25}{40}=0.625 m$$ We can now substitute the values of force, displacement in the work equation, $$W=20Nm \times 0.625 m=12.5 J$$

Therefore, the work done by the force is 12.5 J.

3. A block of mass 10 kg is lifted vertically through a height of 10 m. What is the work done by the force of gravity?

Shortcut Method:

The work done by the force of gravity is given by: $$W = mgh$$ Where,

• (W) is the work done by gravity,
• (m) is the mass of the block,
• (g) is the acceleration due to gravity
• (h) is the height

Substituting the given values, we get:

$$W = 10 * 9.8 * 10 = 980 J$$

Therefore, the work done by the force of gravity is 980 J.

4. *A block of mass 10 kg is moving at a velocity of 5 m/s. The block is brought to