Question

At t = 0, a force F kt is applied on a block making an angle α with the horizontal. Suppose surfaces to be smooth. Find the velocity of the body at the time of breaking off the plane.

• A. $\frac { m g \cos \alpha } { 2 a \sin \alpha }$
• B. $\frac { \mathrm { m } ^ { 2 } \mathrm {~g} ^ { 2 } \cos \alpha } { 2 \mathrm { a } \sin ^ { 2 } \alpha }$
• C. $\frac { \mathrm { mg } ^ { 2 } \cos \alpha } { 2 a \sin ^ { 2 } \alpha }$
• D. $\frac { \mathrm { mg } ^ { 2 } \cos ^ { 2 } \alpha } { 2 \mathrm { sin } \alpha }$

Answer 1

Answer: (C)

$\frac { \mathrm { mg } ^ { 2 } \cos \alpha } { 2 a \sin ^ { 2 } \alpha }$

Answer 2

at cos α = $\frac { \mathrm { mdv } } { \mathrm { dt } }$ or $\mathrm { v } = \frac { \mathrm { at } ^ { 2 } \cos \alpha } { 2 \mathrm {~m} }$

At the break off point mg = at sin α

or t = $\frac { \mathrm { mg } } { \mathrm { a } \sin \alpha }$ .

v = $\frac { a \cos \alpha } { 2 m } \left( \frac { m g } { a \sin \alpha } \right) ^ { 2 } = \frac { m ^ { 2 } \cos \alpha } { 2 a \sin ^ { 2 } \alpha }$ .