Embibe Experts Solutions for Exercise 1: Exercise 5.8

A body moves a distance of $10 \mathrm{m}$ along a straight line under the action of a force of $5 \mathrm{N}$. If the work done is $25 \mathrm{J}$, the angle which the force makes with the direction of motion of the body is.

A force $\left(\frac{m{v}^2}{r}\right)$ is acting on a body of mass $m$ moving with a velocity $v$ in a circle of radius $r.$ What is the work done by the force in moving the body over half the circumference of the circle?

A body of mass $5 \mathrm{kg}$ is moving with a momentum of $10 \mathrm{kg} {\mathrm{ms}}^{-1}$.  A force of $0.2 \mathrm{N}$ acts on it in the direction of motion of the body for $10 \mathrm{s}$. The increase in its kinetic energy is:

A force acts on a 3 g particle in such a way that position of particle as a function of time is given by x = 3t – 4t² + t³, where x is in metre and t is in sec. The work done during first 4s is

A particle moves along the $x$-axis from $x=x_1$ to $x=x_2$ under the influence of a force given by $F=2x.$ Work done in the process is

A $4 \mathrm{kg}$ mass moving with speed $2 \mathrm{m} {\mathrm{s}}^{–1},$ and a $2 \mathrm{kg}$ mass moving with a speed of $4 \mathrm{m} {\mathrm{s}}^{–1},$ are gliding over a horizontal frictionless surface. Both objects encounter the same horizontal force, which directly opposes their motion, and are brought to rest by it. Which statement best describes their respective stopping distances?

A woman lifts a barbell $2.0 \mathrm{m}$ in $5.0 \mathrm{s}.$ If she lifts it by the same distance in $10 \mathrm{s},$ the work done by her is

A particle is moved from $\left(0, 0\right)$ to $\left(a, a\right)$ under a force $\overrightarrow{F}=\left(3\hat{\mathrm{i}}+4\hat{\mathrm{j}}\right)$ from two paths. Path $1$ is $OP$ and path $2$ is $OQP$. Let $W_1$ and $W_2$ be the work done by this force in these two paths. Then