In the arrangement shown in the figure, the ends P and Q of
an unstretchable string move downwards with uniform
speed U. Pulleys A and B are fixed. Mass M moves
upwards with a speed
- $2Ucos\theta$
- $\frac{U}{cos\theta}$
- $\frac{2U}{cos\theta}$
- Ucos$\theta$
The Correct Option is B
Solution and Explanation
$ \, \, \, \, \, t^2=c^2+y^2$
Differentiating this
equation with respect to
time, we ge
$2t\frac{dl}{dt}=0+2y\frac{dy}{dt}$
or $ \, \, \, \, \, \, \, \big(-\frac{dy}{dt}\bigg)=\frac{l}{y}\bigg(-\frac{dl}{dt}\bigg)$
$Here, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, -\frac{dy}{dt}=v_{M} \, \, \, \Rightarrow \, \, \, \frac{1}{y}=\frac{1}{cos\theta}$
and -dl / dt =U
Hence,$ \, \, \, \, \, \, \, \, \, \, v_M=\frac{U}{cos\theta}$
$\therefore \, \, $ Correct option is (b).
Hence ,$ \, \, \, \, \, \, \, \, \, \, v_M=\frac{U}{cos\theta}$
$\therefore \, \, $ Correct option is (b).
Top Questions on Newtons Laws of Motion
Two blocks of masses \( m \) and \( M \), \( (M > m) \), are placed on a frictionless table as shown in figure. A massless spring with spring constant \( k \) is attached with the lower block. If the system is slightly displaced and released then \( \mu = \) coefficient of friction between the two blocks.
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(E) Maximum frictional force can be \( \mu (M + m) g \)Choose the correct answer from the options given below:
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Concepts Used:
Newtons Laws of Motion
Newton’s First Law of Motion:
Newton’s 1st law states that a body at rest or uniform motion will continue to be at rest or uniform motion until and unless a net external force acts on it.
Newton’s Second Law of Motion:
Newton’s 2nd law states that the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the object’s mass.
Mathematically, we express the second law of motion as follows:
Newton’s Third Law of Motion:
Newton’s 3rd law states that there is an equal and opposite reaction for every action.