Solveeit Logo

Question

Physics Question on laws of motion

A block of mass MM is pulled along a horizontal frictionless surface by a rope of mass mm. If a force PP is applied at the free end of the rope, the force exerted by the rope on the block is

A

PmM+m\frac{Pm}{M+m}

B

PmMm\frac{Pm}{M-m}

C

PP

D

PMM+m\frac{PM}{M+m}

Answer

PMM+m\frac{PM}{M+m}

Explanation

Solution

Let acceleration of system (rope + block) is an along the direction of applied force. Then a=PM+ma=\frac{P}{M+m} Draw the FBD of block and rope as shown in figure.where, TT is the required parameter For block T=MaT=Ma \Rightarrow T=MPM+mT=\frac{MP}{M+m}