WORK -

Work:-

In physics, work has a special meaning. Work is only done by force when the force acting on an object produces displacement in it in the direction of force.

Necessary conditions for work done :

(1) Force on an object (2) The object must be displaced (3) Component of force must be along displacement.

E.x- (1) Pushing the wall by hand but the wall does not move, then w=0

(2) A man holding a bucket of water is walking on a horizontal road, then w=0

Work done by constant force:-

The work done on an object by the constant force is defined as the scalar product ( dot product) of the force and the displacement produced.

$\dpi{120} \fn_cm W=\vec{F}.\vec{S}$

$\dpi{120} \fn_cm W=FS\cos\theta$ where $\dpi{120} \fn_cm \theta$ is the angle between $\dpi{120} \fn_cm \vec{F}\;and\;\vec{S}$

It is a scalar quantity and its S.I unit is N-m or Joule. Its CGS unit is Dyne-cm or ergs $\dpi{120} \fn_cm \left [ 1 \;Joule=10^7 \;ergs \right ]$

Special Case:

a. if $\dpi{120} \fn_cm \theta=0^0$

$\dpi{120} \fn_cm W=FS\cos0^0=FS\;(+ve)$

b. if $\dpi{120} \fn_cm \theta=180^0$

$\dpi{120} \fn_cm W=FS\cos180^0=-FS\;(-ve)$

c. $\dpi{120} \fn_cm \theta=90^0$

$\dpi{120} \fn_cm W=FS\cos90^0=0$

NOTE:-

(a) Between $\dpi{120} \fn_cm 0^0\leq \theta< 90^0$ work is +ve and between $\dpi{120} \fn_cm 90^0<\theta\leq 180^0$ work is -ve and at $\dpi{120} \fn_cm \theta=90^0\;or\;F=0\;or\;S=0$ work is zero

(b) Total work done is equal to the sum of work done by individual forces on body/ system.

(c) Weight of body=50 kgf that means m=50 kg.

(d) -ve work means velocity decreases and +ve work means velocity increases.

(e) If a force is always tangential to the path and its magnitude is constant then work is given by W=±FS where S is the path length (distance)

Ex-

$\dpi{120} \fn_cm dw=Fds\cos0^0$

$\dpi{120} \fn_cm w=F\int ds=F\left ( \frac{\pi r}{2} \right )$

Exercise 1

1. How much work is done in pulling a box 2m across a tabletop with a horizontal force of 35N? (70 J)

2. A woman pushes a lawn roller with a force of 180N at an angle of 37° downward from the horizontal. How much work does she do as she pushes the roller a horizontal distance of 50m?

3. A worker does 300 J of work against a frictional retarding force of 15N in pushing a box across a floor in 3 sec. If the box moves with a constant speed, how fast is it going? (6.7 m/s)

4. Find the work done in moving a 50 kg block through a horizontal distance of 10m by applying a force of 100N which makes an angle of 60° with the horizontal. (500 J)

5. A coin of mass m slides a distance D along a tabletop. If the coefficient of friction between the coin and the table is μ, Find the work done on the coin by friction. ( -μmgD)

6. The coefficient of kinetic friction between a 20 kg box and the floor is 0.4. How much work does a pulling force do on the box in pulling it 8m across the floor at a constant speed? The pulling force is directed 37° above the horizontal. (482 J)

7. A 52 kg box rests on the floor. How much work is required to move it at constant speed (a) 12 m along the floor against a frictional force of 150N and (b) 12 m vertically upward? (1.8×10³ J, 6.9×10³ J)

8. How much work is needed to push a 1250 kg car 10 m up a 37 incline at constant velocity? (a) Ignore friction (b) assume the coefficient of kinetic friction is 0.09.

9. A particle moves from position $\dpi{120} \fn_cm (3\hat{i}+2\hat{j}-6\hat{k})$ to position $\dpi{120} \fn_cm (14\hat{i}+13\hat{j}+9\hat{k})$ in meters and a constant force $\dpi{120} \fn_cm (4\hat{i}+2\hat{j}+3\hat{k})$ N acts on it. Calculate the work done by the force. (111 N-m)

10. A man cycles up a hill whose slope is 1 in 20. If the mass of the man and bicycle is 150 kg and the man covers a distance of 100m along the incline, calculate the work done by man. (7350 J)

11. A particle moves 4m along z-axis. Find the workdone by the force $\dpi{120} \fn_cm (-\hat{i}+2\hat{j}+3\hat{k})$ N. (12J)

Exercise

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