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The work done in pulling up a block of wood weighing 2 kN for a length of 10m on a smooth plane inclined at an angle of 15° with the horizontal is (given: sin15°=0.2588):

4.36 kJ

5.17 kJ

8.91 kJ

9.82 kJ

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A force $\vec{F} = 5 \hat{i} + 6 \hat{j} - 4 \hat{k}$ acting on a body, produces a displacement $\vec{s} = 6 \vec{i} + 5 \vec{k} .$ Work done by the force is

18 units

15 units

12 units

10 units

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Explanation

$W = \vec{F} . \vec{s} = (5 \hat{i} + 6 \hat{j} - 4 \hat{k}) . (6 \hat{i} + 5 \hat{k}) = 30 - 20 = 10$ units

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A body of mass 6kg is under a force which causes displacement in it given by $S = \frac{t^{2}}{4}$ metres where t is time. The work done by the force in 2 seconds is-

12 J

9 J

6 J

3 J

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Explanation

$s = \frac{t^{2}}{4}$ ∴ $d s = \frac{t}{2} d t$

$F = m a = \frac{m d^{2} s}{d t^{2}} = \frac{6 d^{2}}{d t^{2}} [\frac{t^{2}}{4}] = 3 N$

Now $W = \int_{0}^{2} F d s = \int_{0}^{2} 3 \frac{t}{2} d t = \frac{3}{2} {[\frac{t^{2}}{2}]}_{0}^{2} = \frac{3}{4} [{(2)}^{2} - {(0)}^{2}] = 3 J$

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A force of $(3 \hat{i} + 4 \hat{j})$ Newton acts on a body and displaces it by $(3 \hat{i} + 4 \hat{j}) m .$ The work done by the force is

10 J

12 J

16 J

25 J

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Explanation

$W = \vec{F} . \vec{s}$

$= (3 \hat{i} + 4 \hat{j}) . (3 \hat{i} + 4 \hat{j}) = 9 + 16 = 25 J$

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A force $\vec{F} = 6 \hat{i} + 2 \hat{j} - 3 \hat{k}$ acts on a particle and produces a displacement of $\vec{s} = 2 \hat{i} - 3 \hat{j} + x \hat{k} .$ If the work done is zero, the value of x is

– 2

1/2

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Explanation

$W = \vec{F} . \vec{s} = (6 \hat{i} + 2 \hat{j} - 3 \hat{k}) . (2 \hat{i} - 3 \hat{j} + x \hat{k}) = 0$

$12 - 6 - 3 x = 0$ ⇒ $x = 2$

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A particle moves from position ${\vec{r}}_{1} = 3 \hat{i} + 2 \hat{j} - 6 \hat{k}$ to position ${\vec{r}}_{2} = 14 \hat{i} + 13 \hat{j} + 9 \hat{k}$ under the action of force $4 \hat{i} + \hat{j} + 3 \hat{k} N .$ The work done will be

100 J

50 J

200 J

75 J

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Explanation

Work done = force x displacement

$W = \vec{F} . (\vec{r_{2}} - \vec{r_{1}}) = (4 \hat{i} + \hat{j} + 3 \hat{k}) (11 \hat{i} + 11 \hat{j} + 15 \hat{k})$

$W = 44 + 11 + 45 = 100 J o u l e$

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A force $(\vec{F}) = 3 \hat{i} + c \hat{j} + 2 \hat{k}$ acting on a particle causes a displacement: $(\vec{s}) = - 4 \hat{i} + 2 \hat{j} + 3 \hat{k}$ in its own direction. If the work done is 6 J then the value of ‘c’ is

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Explanation

$W = (3 \hat{i} + c \hat{j} + 2 \hat{k}) . (- 4 \hat{i} + 2 \hat{j} + 3 \hat{k}) = 6 J o u l e$

$W = - 12 + 2 c + 6 = 6$ ⇒ $c = 6$

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A force $\vec{F} = (5 \hat{i} + 4 \hat{j})$ N acts on a body and produces a displacement $\vec{S} = (6 \hat{i} - 5 \hat{j} + 3 \hat{k})$ m. The work done will be

10 J

20 J

30 J

40 J

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Explanation

$W = \vec{F} . \vec{s} = (5 \hat{i} + 4 \hat{j}) . (6 \hat{i} - 5 \hat{j} + 3 \hat{k})$

$= 30 - 20 = 10 J$

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A uniform chain of length 2m is kept on a table such that a length of 60cm hangs freely from the edge of the table. The total mass of the chain is 4kg. What is the work done in pulling the entire chain on the table

7.2 J

3.6 J

120 J

1200 J

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Explanation

$M a s s o f t h e h a n g i n g l e n g t h = \frac{60}{200} \times 4 = 1.2 k g W o r k d o n e i n p u l l i n g t h e c h a i n = p o t e n t i a l e n e r g y o f t h e h a n g i n g c h a i n = m g h = 1.2 \times 10 \times \frac{60 \times 10^{- 2}}{2} = 3.6 J$

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A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle, the motion of the particle takes place in a plane. It follows that

Its velocity is constant

Its acceleration is constant

Its kinetic energy is constant

It moves in a straight line

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Explanation

When a force of constant magnitude which is perpendicular to the velocity of particle acts on a particle, work done is zero and hence change in kinetic energy is zero.

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