Question

A system of two blocks of masses m = 2 kg and M = 8 kg is placed on a smooth table as shown in the figure. The coefficient of static friction between two blocks is 0.5. The maximum horizontal force F that can be applied to the block of mass M so that the blocks move together will be

• A. 9.8 N
• B. 39.2 N
• C. 49 N
• D. 78.4 N

Answer 1

The Correct Option is C

To determine the maximum horizontal force \(F\) that can be applied to the block of mass \(M\) such that the blocks move together, we need to consider the force of static friction between the two blocks. The blocks will move together as long as the force of static friction is not exceeded.

The formula for the force of static friction \(f_s\) is:

\(f_s = \mu_s \times N\)

where \(\mu_s = 0.5\) is the coefficient of static friction and \(N\) is the normal force. Since the block \(m\) is resting on the block \(M\), the normal force is equal to the weight of the block \(m\).

The weight of block \(m\) is given by:

\(N = m \times g\)

where \(m = 2\ \mathrm{kg}\) and \(g = 9.8\ \mathrm{m/s^2}\).

Calculating the normal force:

\(N = 2 \times 9.8 = 19.6\ \mathrm{N}\)

Now, calculate the maximum static friction:

\(f_s = 0.5 \times 19.6 = 9.8\ \mathrm{N}\)

For the two blocks to move together, the maximum force \(F_{\text{max}}\) is the force that causes an acceleration equal to the static friction as it acts on both blocks:

\(F_{\text{max}} = (m + M) \times a\)

Since the acceleration \(a\) is equal to the static friction divided by the mass \(m\):

\(a = \frac{f_s}{m} = \frac{9.8}{2} = 4.9\ \mathrm{m/s^2}\)

Now, compute \(F_{\text{max}}\):

\(F_{\text{max}} = (2 + 8) \times 4.9 = 10 \times 4.9 = 49\ \mathrm{N}\)

Therefore, the maximum horizontal force \(F\) that can be applied is 49 N.