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

Question

A block of mass M is pulled along a horizontal frictionless surface by a rope of mass m. If a force P is applied at the free end of the rope, the force exerted by the rope on the block is (A)  (Pm/M+m)  (B)  (Pm/M-m)  (C)  P  (D)  (PM/M+m)

Tardigrade Admin · Accepted Answer

The acceleration of the (rope + block) system is given by a=(P/m+M). Hence, the force on the block of mass M is equal to ((M P/m+M))

Q. A block of mass $M$ is pulled along a horizontal frictionless surface by a rope of mass $m$ . If a force $P$ is applied at the free end of the rope, the force exerted by the rope on the block is

$\frac{P m}{M + m}$

$\frac{P m}{M - m}$

$P$

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

The acceleration of the (rope + block) system is given by $a = \frac{P}{m + M}$ .
Hence, the force on the block of mass $M$ is equal to $(\frac{MP}{m + M})$

Q. A block of mass M is pulled along a horizontal frictionless surface by a rope of mass m. If a force P is applied at the free end of the rope, the force exerted by the rope on the block is

M+mPm​

M−mPm​

P

M+mPM​

The acceleration of the (rope + block) system is given by a=m+MP​. Hence, the force on the block of mass M is equal to (m+MMP​)

Q. A block of mass $M$ is pulled along a horizontal frictionless surface by a rope of mass $m$ . If a force $P$ is applied at the free end of the rope, the force exerted by the rope on the block is

$\frac{P m}{M + m}$

$\frac{P m}{M - m}$

$P$

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

The acceleration of the (rope + block) system is given by $a = \frac{P}{m + M}$ .
Hence, the force on the block of mass $M$ is equal to $(\frac{MP}{m + M})$