Problem 1. A body of mass 5 kg is placed at the origin, and can move only on the x-axis. A force of 10 N is acting on it in a direction making an angle of 600 with the x-axis and displaces it along the x-axis by 4 metres. The work done by the force is
(a)2.5 J (b)7.25 J (c)40 J (d)20 J
Solution : (d) Work done \( = \overrightarrow F .\,\overrightarrow s = Fs\cos \theta = 10 \times 4 \times \cos 60^0 = 20J \)
Problem 2 : A force \( F = (5\mathop i\limits^ \wedge + 3\mathop j\limits^ \wedge )N \) is applied over a particle which displaces it from its origin to the point \( r = (2\mathop i\limits^ \wedge - 1\mathop j\limits^ \wedge ) \) metres. The work done on the particle is
(a)–7 J (b)+13 J (c)+7 J (d)+11 J
Solution : (c) Work done \( = \overrightarrow F .\,\overrightarrow r = (5i + 3\mathop j\limits^ \wedge )\,.\,(2\mathop i\limits^ \wedge - 1\mathop j\limits^ \wedge ) \) =10 - 3 = +7J
Problem 3. A horizontal force of 5 N is required to maintain a velocity of 2 m/s for a block of 10 kg mass sliding over a rough surface. The work done by this force in one minute is
(a)600 J (b)60 J (c)6 J (d)6000 J
Solution : (a) Work done = Force x displacement =
F x s = F x v x t = 5 x 2 x 60 = 60 = 600J
Problem 4. A box of mass 1 kg is pulled on a horizontal plane of length 1 m by a force of 8 N then it is raised vertically to a height of 2m, the net work done is
(a)28 J (b)8 J (c)18 J (d)None of above
Solution : (a) Work done to displace it horizontally = F x s = 8 x 1 = 8J
Work done to raise it vertically F ´s = mgh = 1 ´ 10 ´ 2 = 20 J
Net work done = 8 +20 = 28 J
Problem 5. A 10 kg satellite completes one revolution around the earth at a height of 100 km in 108 minutes. The work done by the gravitational force of earth will be
(a) \(
108 \times 100 \times 10\smallint
\) (b) \(
\frac{{108 \times 10}}
{{100}}\smallint
\) (c) \(
\frac{{100 \times 10}}
{{108}}\smallint
\) (d)Zero
Solution : (d) Work done by centripetal force in circular motion is always equal to zero.