Consider a block and trolley system as shown in figure. If the coefficient of kinetic friction between the trolley and the surface is 0.04, the acceleration of the system in m/s2 is : (Consider that the string is massless and unstretchable and the pulley is also massless and frictionless) :
The kinetic frictional force is given by:
\[ f_k = \mu N = 0.04 \times 20g = 8 \, \text{Newton} \]
The net force acting on the system is:
\[ F_{\text{net}} = 60 \, \text{N} - f_k = 60 - 8 = 52 \, \text{N} \]
The total mass of the system is:
\[ m_{\text{total}} = 20 \, \text{kg} + 6 \, \text{kg} = 26 \, \text{kg} \]
The acceleration of the system is:
\[ a = \frac{F_{\text{net}}}{m_{\text{total}}} = \frac{52}{26} = 2 \, \text{ms}^{-2} \]
A wooden block of mass M lies on a rough floor. Another wooden block of the same mass is hanging from the point O through strings as shown in the figure. To achieve equilibrium, the coefficient of static friction between the block on the floor and the floor itself is
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: