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Acceleration Of Block On Horizontal Smooth Surface MCQ - Practice Questions with Answers
Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET
Quick Facts
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6 Questions around this concept.
Solve by difficulty
A force of 50N acts in the direction as shown in fig. The block of mass of 5kg resting on a smooth horizontal surface . What is the acceleration of the block?
In fig, if the surfaces are frictionless then tension $T_2$ (in N ) will be:
Two blocks of mass 5 kg and 5 kg are connected by a string of mass 2 kg as shown in fig. Point Q is the midpoint of the string. A pulling force of 24 N is applied on block 5 kg at an angle of $60^{\circ}$ with horizontal. The tension (in N ) at point Q is (all surfaces are frictionless).
At time $t=0$, a force $F=k t$ where k is any constant, is exerted on a small body with mass $m$, which is initially at rest on a frictionless horizontal surface. The resultant direction of the applied force forms an angle of $90^{\circ}$ with the horizontal plane. Find the velocity of the body at the moment of its breaking off plane.
A horizontal force of 10 N is applied to block A as shown in the figure. The mass of blocks A and B are 2 kg and 3 kg respectively. The blocks slide over a frictionless surface. The force exerted by block A on block B is :
Concepts Covered - 1
Acceleration of Block on horizontal smooth surface
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Free Body Diagram (FBD)
In this diagram, the object of interest is isolated from its surroundings and the interactions between the object and the surroundings are represented in terms of forces.
After drawing FBD, Choose the axes and write the equation of motion. This is very helpful while solving questions.
Example-
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When the pull is horizontal, and no friction
Balance forces
the body is moving in the x-axis
$\begin{aligned} & \because F_y=0 \\ & R=m g \quad \& \quad F=m a \\ & a=\frac{F}{m}\end{aligned}$
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Pull Acting at Angle (Upward)
Balancing forces in both X and Y direction,
$\begin{aligned} & R+F \sin \theta=m g \\ & R=m g-F \sin \theta \text { along } Y \text {-axis } \\ & F \cos \theta=m a \text { along } X \text {-axis } \\ & a=\frac{F \cos \theta}{m}\end{aligned}$
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Push Acting at Angle (Downward)
Balancing forces in both X and Y direction,
$\begin{aligned} & R=m g+F \sin \theta \text { along } Y \text {-axis } \\ & a=\frac{F \cos \theta}{m}_{\text {along } X \text {-axis }}\end{aligned}$
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