Two block of masses M(1) and, M(2) are c...

Two block of masses `M_(1)` and, `M_(2)` are connected with a string which passed over a smooth pulley . The mass `M_(1)` is placed on a tough inclined plane as shown in figure .The coefficient of friction between and block and the inclined plane is `mu` what should be the minimum mass `M_(2)` so that the block `M_(1)` slides upwards?

`M_(2) = M_(1) (sin theta + mu cos theta)`

`M_(2) = M_(1) (sin theta - mu cos theta)`

`M_(2) = (M_(1))/(sin theta + mu cos theta)`

`M_(2) = (M_(1))/(sin theta - mu cos theta)`

Verified by Experts

The correct Answer is:

For upward accelerationof `M_(1)`:
`M_(2)g geM_(1) g sin theta + mu M_(1) g cos theta`
`rArr (M_(2))_(max) = M_(1) (sin theta + mu cos theta)`

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Two blocks of masses m_(1) and m_(2) are connected by a string of negligible mass which pass over a frictionless pulley fixed on the top of an inclined plane as shown in figure. The coefficient of friction between m_(1) and plane is mu .

If `m_(1)=m_(2)`, the mass `m_(1)` first begin to move up inclined plane when the angle of inclination `theta`, then `mu=sec theta-tan theta`

If `m_(1)=m_(2)`, then mass `m_(1)` first begin to side down the plane if `mu=sec theta-tan theta`.

If `m_(1)=2m_(2)`, then mass `m_(1)` first begins to slide down the plane if `mu=2 tan theta`.

If `m_(1)=2m_(2)`, then mass `m_(1)` first begins to slide down the plane if `mu= tan theta - 1/2 sec theta`.

static in nature in the direction up the plane and have the value `30.2N`

static in nature in the direction down the plane and have the value `30.2N`

kinetic in nature in the direction up the plane and have the value `30.2N`

None of these

`8 m//s^(2)`

`4m//s^(2)`

`6m//s^(2)`

`3 m//s^(2)`