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

Let acceleration of system (rope + block) is a along the direction of applied force. Then,

a = \frac{P}{M + m}

Draw the FBD of block and rope as shown in figure.
where, T is the required parameter. For block T=Ma

\Rightarrow

T = \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

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

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

P

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

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

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​

Let acceleration of system (rope + block) is a along the direction of applied force. Then, a=M+mP​ Draw the FBD of block and rope as shown in figure.where, T is the required parameter. For block T=Ma ⇒ T=M+mMP​

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

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

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

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