Question

Two blocks, each having mass M, rest on frictionless surface as shown in the fig. If the pulleys are light and frictionless and M on the incline is allowed to move, then the tension in the string will be? E 11th question

Answer 1

Answer:

The tension in the string will be $\bold{\frac{m g \sin \theta}{2}}$.

Explanation:

In the given diagram the mass that is placed on inclination has two tensions working one due to gravitational pull on the weight and the other on the string.

So, $m g \sin \theta-T=m a$ is the force acting on the mass that is allowed to slide let this be equation 1

The other mass is not moving and is static so there will be only one force acting due to tension

T = ma is the equation for second mass

Subtracting equation 2 from 1,

$2 T=m g \sin \theta$

$\bold{T=\frac{m g \sin \theta}{2}}$

Answer 2

This might help(always draw a diagram first)