Q: A force of 7 N acts on an object. The displacement is 8 m, in the direction of the force. Consider force acting on the object through the displacement. What is the work done in this case ?

\$Given, \: force =7 \: N \;; \: displacement = 8 \: m \$\$Work \: done = Force \times Displacement\$ \$= 7 \times 8 = 56 \: J\$

Q: A rigid body moves a distance of 10m along a straight line under the action of a force of 5 N. lf the work done by this force on the body is 25 joules, the angle which the force makes with the direction of motion of the body is

we know,\$W=FS cos\theta\$\$cos\theta=\dfrac{W}{FS}\$\$cos\theta=\dfrac{25}{50}\$\$cos\theta=\dfrac{1}{2}\$\$\theta=cos^{-1}0.5\$\$\theta=60^0\$

Q: A man pushes a wall and fails to displace it. He does :

\$\textbf{Hint}\$: We know work done in displacing the object is equal to the product of the force applied on the object and the distance traveled by it in the direction of force applied. Let us apply this concept to solve the problem. \$\textbf{Step 1}\$: We know \$Work \quad done(W)=F.S\$This is the work done by the man when he pushes the wall. Given that man pushes a wall and fails to displace it so displacement is zero i.e.., \$S=0\$\$\textbf{Step 2}\$: \$Work \quad done(W)=F.0=0\$So work done is zero. Thus option (D) is correct.

Question 1

Convert $1$ joule to ergs.

Accepted Answer

Joule : SI system, erg:CGS systemWork = Force x Distance = Mass x Acceleration x Length = Mass x $\dfrac{Length}{(Time)^2} \times Length$Dimensions of work = $[W] = [M^1L^2T^{-2}]$$\therefore a = 1, b = 2, c = -2.$NowSI system&#9;     CGS system$M_1 = 1 kg$&#9;     $M_2 = 1 kg$$L_1 = 1 m$&#9;     $L_2 = 1 cm$$T_1 = 1s$&#9;     $T_2 = 1 s$Here $N_1 = 1, N_2 = ?$$\therefore Using\,\, N_2 = N_1\left [\dfrac{M_1}{M_2}\right]^a \left[\dfrac{L_1}{L_2}\right]^b \left[\dfrac{T_1}{T_2}\right]^c$$= 1\left[\dfrac{1kg}{1g}\right]^1 \left [\dfrac{1m}{1 cm}\right]^2 \left[\dfrac{1 s}{1 s}\right]^{-2}$$= 1\left[\dfrac{1000 g}{1 g}\right]^1 \left[\dfrac{100 cm}{1 cm}\right]^2$$= 10^7$$\therefore N_2 = 10^7$So $1 J = 10^7$ erg

Question 2

A force of 7 N acts on an object. The displacement is 8 m, in the direction of the force. Consider force acting on the object through the displacement. What is the work done in this case ?

Accepted Answer

$Given, \: force =7 \: N \;; \: displacement = 8 \: m $$Work \: done = Force \times  Displacement$                      $=  7 \times  8 = 56 \: J$

Question 3

A rigid body moves a distance of 10m along a straight line under the action of a force of 5 N. lf the work done by this force on the body is 25 joules, the angle which the force makes with the direction of motion of the body is

Accepted Answer

we know,$W=FS cos\theta$$cos\theta=\dfrac{W}{FS}$$cos\theta=\dfrac{25}{50}$$cos\theta=\dfrac{1}{2}$$\theta=cos^{-1}0.5$$\theta=60^0$

Question 4

Explain Positive, Negative and Zero Work. Give one example of each.

Accepted Answer

POSITIVE WORK: The work done on an object is said to be positive work when force and displacement are in same direction.Example: When an object moves on horizontal surface, force and displacement acts in same direction. So, work done is positive.NEGATIVE WORK: The work done is said to be negative work when force and displacement are in opposite direction.Example: When an object is thrown upwards,the force of gravity is in downward direction whereas displacement acts in upward direction. ZERO WORK: The work done is said to be zero when force and displacement are perpendicular to each other or when either force or displacement is zero.Example: When we hold an object and walk, the force acts in downward direction whereas displacement acts in forward direction.

Question 5

Give two examples of (i) positive work (ii) negative work (iii) zero work

Accepted Answer

Positive work -1. Pushing a box through a distance.2. Lift a book through a height.Negative work -1. Work done by force of gravity on an object when lifted.2. Work done by friction when we walk.Zero work -1. Pushing a wall.2. Walking while holding an object.

Question 6

A cyclist comes to a skidding stop in $10 m$ . During this process, the force on the cycle due to the road is $200 N$ and is directly oppose to the motion.
(a) How much work does the road do on the cycle?
(b) How much work does the cycle do on the road?

Accepted Answer

(a) Given Force on the cycle due to the road= 200 N and distance traveled during skidding before stopping =10 mWork done = force$\times$displacement=200$\times$10=$-2000 J$ (negative sign used as applied in the direction opposite to that of motion   (b) zero

Question 7

A man pushes a wall and fails to displace it. He does :

Accepted Answer

$\textbf{Hint}$: We know work done in displacing the object is equal to the product &#10;of the force applied on the object and the distance traveled by it in &#10;the direction of force applied. Let us apply this concept to solve the problem. $\textbf{Step 1}$: We know $Work \quad done(W)=F.S$This is the work done by the man when he pushes the wall. Given that man pushes a wall and fails to displace it so displacement is zero i.e.., $S=0$$\textbf{Step 2}$: $Work \quad done(W)=F.0=0$So work done is zero. Thus option (D) is correct.

Question 8

A block of mass $m = 2.50  kg$ is pushed a distance $d  =  2.20  m$ along a frictionless, horizontal table by a constant applied force of magnitude $F  = 16.0 N$ directed at an angle $\theta = 25.0^{0}$ below the horizontal as shown in above figure. Determine the work done on the block by (a) the applied force, (b) the normal force exerted by the table, (c) the gravitational force, and (d) the net force on the block.

Accepted Answer

We apply the definition of work by a constant force in the first three parts, but then in the fourth part we add up the answers. The total (net) work is the sum of the amounts of work done by the individual forces, and is the work done by the total (net) force. This identification is not represented by an equation in the chapter text, but is something you know by thinking about it, without relying on an equation in a list.The definition of work by a constant force is $W = F&#916;r \cos&#952;$.(a) The applied force does work given by$W = F&#916;r \cos&#952; = (16.0 N)(2.20 m)\cos25.0^{0} = 31.9 J$(b), (c) The normal force and the weight are both at $90^{0}$ to the displacement in any time interval. Both do $0$ work.(d) $\sum W = 31.9 J + 0 + 0 = 31.9 J$

Question 9

A cyclist comes to skidding stop in 10m. During this process, the force on the cycle due to the road is 200 N and is directly opposed to the motion.(a) how much work does the road do on the cycle? (b) How much work does the cycle do on the road?

Accepted Answer

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Question 10

A mass of $10\ kg$ is at a point $A$ on a table. It is moved to a point $B$. If the line joining $A$ and $B$ is horizontal what is the work done on the object by the gravitational force? Explain your answer.

Accepted Answer

Hint: Observe the angle between force of gravity and displacement.Explanation:Step 1: Concept used:Work done is given by:$W= \overrightarrow {F}. \overrightarrow {S}$$W=F\ S\cos\theta$where, $F = $ force applied$S = $ displacement$\theta = $ angle between the force applied and displacement.Step 2: Finding the work done:In the given problem, the force of gravity and displacement are perpendicular to each other. $\theta =$ $90$$ \therefore W=0$ Hence, the work done is zero.

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