Question

Two blocks are arranged at the ends of a massless cord over a frictionless massless pulley as shown in the figure. Assume the system starts from rest. When the masses have moved a distance of 0.355 m, their speed is 1.32 m/s. The acceleration of gravity is 9.8 m/s 2 . 4.1 kg 2.7 kg µ What is the coefficient of friction between m2 and the table?

Answer 1

Answer:

The coefficient of friction is 0.242.

Explanation:

Given

Mass on the table is, $m_1 = 4.1$ kg

Hanging mass is, $m_2 = 2.7$ kg

Displacement of the masses is, $s = 0.355$ m

Initial velocity of the masses is, $u=0$ m/s

Final velocity of the masses is, $v = 1.32$ m/s

Acceleration by gravity is $g = 9.8$ m/s².

The acceleration of the system is calculated using equation of motion and is given as:

$a=\frac{v^2-u^2}{2s}\\a=\frac{1.32^2-0}{2\times 0.355}=2.46\ m/s^2$

Therefore, $a = 2.46$ m/s²

Applying Newton's second law on the hanging mass, we get:

$m_2\times g-T = m_2\times a\\T = m_2\times (g - a)\\T=2.7\times(9.8 - 2.46)\\T = 19.82\ N$

Now, applying Newton's second law on the mass on table, we get:

$T - F_f = m_1\times a\\F_f =T - m_1\times a\\F_f = 19.82 -4.1\times 2.46\\F_f = 9.73\ N$

The normal force of the table on the mass is given as:

$N=m_1\times g=4.1\times 9.8=40.18\ N$

The coefficient of friction is the ratio of the frictional force and the normal force and is given as:

$\mu=\frac{F_f}{N}=\frac{9.73}{40.18}=0.242$

Therefore, the coefficient of friction between the mass on the table and the table is 0.242.