A block of mass 0.50 kg is on a table and attached to a block of mass 0.25 kg by a taut,massless string.The 0.25-kg block hangs by means of the string,which goes over a pulley.The 0.25-kg block is released,and it descends from rest at a constant acceleration,travelling 1.00 m in 1.25 s.What is the coefficient of kinetic friction between the table and the 0.50-kg block?
A)0.30
B)0.47
C)0.53
D)0.80

Question

Quizplus · Accepted Answer

The acceleration of the system can be calculated using the formula $ a = \frac{2d}{t^2} $, which gives $ a = \frac{2 \times 1.00}{1.25^2} = 1.28 \, \text{m/s}^2 $. Using Newton's second law and considering the forces acting on both blocks, the coefficient of kinetic friction $ \mu_k $ can be found using the equation $ \mu_k = \frac{m_2g - m_2a - m_1a}{m_1g} $, which simplifies to approximately 0.30.

A Block of Mass 0