First slide

Second law

Question

A block of mass M is pulled along horizontal frictionless surface by a rope of mass m. Force p is applied at one end of rope Fig. The force which
the rope exerts on the block is

Easy

A

$\frac{P}{(M - m)}$
B

$\frac{P}{M (m + M)}$
C

$\frac{PM}{(m + M)}$
D

$\frac{PM}{(M - m)}$

Solution

Let o be the common acceleration of the system. Here
T =M o for block
P -T =m o for rope
$∴$ P - M a=ma or P = a(M+m)
or $a = \frac{P}{(M + m)}$
Now $T = \frac{MP}{(M + m)}$

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