Basics of Statics

A rope of length $8 \mathrm{m}$ and linear density $0.5 \mathrm{kg} {\mathrm{m}}^{-1}$ is lying lengthwise on a horizontal smooth floor. It is pulled by a force of $12 \mathrm{N}.$ The tension at the mid-point would be:-

A block of mass $m$ is suspended by a light thread from an elevator. The elevator is accelerating upward with uniform acceleration $a$. The work done by tension on the block during $t$ seconds is $(u=0):$

The minimum acceleration with which a fireman can slide down a rope whose breaking strength is two third of his weight?

A uniform string of length $10 \mathrm{m}$ & mass $20 \mathrm{kg}$ lies on a smooth frictionless inclined plane. A force of $200 \mathrm{N}$ is applied as shown. The tension at the mid point of string is

A man of mass $m$ stands on a crate of mass $M$. He pulls on a light rope passing over a smooth light pulley. The other end of the rope is attached to the crate. For the system to be is equilibrium, the force exerted by the man on the rope will be

Two blocks of masses $m_1=4$ kg and $m_2=2$ kg are connected to the ends of a string which passes over a massless, frictionless pulley. The total downwards thrust on the pulley is nearly

In the figure shown, $m_1=10 \mathrm{k}\mathrm{g},\mathrm{ }{ m}_2=6 \mathrm{k}\mathrm{g},\mathrm{ }\mathrm{ }{m}_3=4 \mathrm{k}\mathrm{g}.$ If $T_3=40 \mathrm{N},\mathrm{ }\mathrm{ }{T}_2=?$

A mass of $10 \mathrm{kg}$ is suspended from a string balance. It is pulled aside by a horizontal string so that it makes an angle of $60°$ with the vertical. The new reading of the balance is

Assertion: string can never remain horizontal, when loaded at the middle, howsoever large the tension may be.

Reason: For horizontal string, angle with vertical, $\theta =90°, T = \frac{W}{2cos\theta } = \frac{W}{2cos90°} =\infty$ .

A pendulum having a bob of mass $m$ is hanging in a ship sailing along the equator form east to west. When the ship is stationary with respect to water, the tension in the string is $T_0$ . The difference between $T_0$ and earth attraction on the bob, is

A body of mass $2 \mathrm{kg}$ moving on horizontal surface with an initial velocity of $4 {\mathrm{ms}}^{-1}$ comes to rest after $2 \mathrm{s}$. If one wants to keep this, body moving on the same surface with a velocity $4 {\mathrm{ms}}^{-1}$, the force required is

A body of mass $2 \mathrm{kg}$ moving on horizontal surface with an initial velocity of $4 {\mathrm{ms}}^{-1}$ comes to rest after $2 \mathrm{s}$. If one wants to keep this, body moving on the same surface with a velocity $4 {\mathrm{ms}}^{-1}$, the force required is

A body of $0.5 \mathrm{kg}$ moves along the positive $x$-axis under the influence of a varying force $F$ (in newton) as shown below.

If the speed of the object at $x=4 \mathrm{m}$ is $3.16 \mathrm{m} {\mathrm{s}}^{-1}$ then its speed at $x=8 \mathrm{m}$ is

The rough surface of a horizontal table offers a definite maximum opposing force to initiate the motion of a block along the table, which is proportional to the resultant normal force given by the table. Forces $\mathrm{F}1$ and $\mathrm{F}2$ act at the same angle $\mathrm{\theta }$ with the horizontal and both are just initiating the sliding motion of the block along the table. Force $\mathrm{F}1$ is a pulling force while the force $\mathrm{F}2$ is a pushing force. $\mathrm{F}2 > \mathrm{F} 1$ , because

A constant force acting on a body of mass of $5 \mathrm{kg}$ changes its speed from $5 \mathrm{m}\mathrm{ }{\mathrm{s}}^{-1}\mathrm{ }$to $10 \mathrm{m} {\mathrm{s}}^{-1}$ in $10 \mathrm{s}$ without changing the direction of motion. What is the force acting on the body?

Let the forces acting on a particle of mass $4 \mathrm{kg}$ for $t<0,\mathrm{ }0<t<4 \mathrm{s},\mathrm{ }t>4 \mathrm{s}$ be $F_1, F_2$ and $F_3$ respectively. Which of the following statements is true regarding the position-time graph of the particle given below?

A body of mass 6 kg moves in a straight line according to the equation,
$x=t^3-75t$,
where $x$ denotes the distance in metre and $t$ the time in second. The force on the body at $t=4$ s is

A force of $100 \mathrm{dyne}$ acts a mass of $5 \mathrm{gram}$ for $10 \mathrm{s}$. The velocity produced is