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.

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 = FΔr \cosθ\$.(a) The applied force does work given by\$W = FΔr \cosθ = (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\$

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Nature of Work Done

Q: Convert \$1\$ joule to ergs.

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 CGS system\$M_1 = 1 kg\$ \$M_2 = 1 kg\$\$L_1 = 1 m\$ \$L_2 = 1 cm\$\$T_1 = 1s\$ \$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

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

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.

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

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.

Q: 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?

(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

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.

Q: 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.

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.

Physics

definition

Condition for Positive, Negative & Zero work done by a force

We know that:

W = F . d = F d c o s θ

Sign of work done depends on the angle between force and displacement vector.
If

θ = 0^{0}

- Work done is positive.
If

θ = 9 0^{0}

- Work done is zero.
If

θ = 18 0^{0}

- Work done is negative.

definition

Positive, Negative and Zero work done

1. Positive Work : If a force displaces the object in its direction, then the work done is positive. The example of this kind of work done is motion of ball falling towards ground where displacement of ball is in the direction of force of gravity.

2. Negative Work : If the force and the displacement are in opposite directions, then the work is said to be negative. For example if a ball is thrown in upwards direction, its displacement would be in upwards direction but the force due to earths gravity is in the downward direction.

3. Zero Work : If the directions of force and the displacement are perpendicular to each other, the work done by the force on the object is zero.
For example, when we push hard against a wall, the force we are exerting on the wall does no work, because in this case the displacement of the wall is d = 0. However, in this process, our muscles are using our internal energy and as a result we get tired.

definition

Sign convention

Work done by the system i.e. work done by the gas is positive. Example: expansion of gas on heating. Work done on the system is negative. Example: A constant force being applied on the piston of a cylinder.

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Important Questions

Convert $1$ joule to ergs.

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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 ?

Medium

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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

Medium

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

Medium

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

Medium

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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?

Medium

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

Medium

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A block of mass $m = 2.50 k g$ 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 $θ = 25 . 0^{0}$ 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.

Medium

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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?

Medium

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A mass of $10 k g$ 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.

Medium

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definition

Condition for Positive, Negative & Zero work done by a force

definition

Positive, Negative and Zero work done

definition

Sign convention

REVISE WITH CONCEPTS

What is Work?

Work and Work Done by a Constant Force

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Basics of Work Done

Convert 1 joule to ergs.

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 ?

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

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

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

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?

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

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?

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.

Convert $1$ joule to ergs.

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?

A mass of $10 k g$ 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.