Work, Power and EnergyMedium
Question
A mass of 1 kg kept in a inclined plane with $30^{\circ}$ inclination with respect to horizontal plane and it is at rest initially. Then the whole assembly is moved up with constant velocity of $4\text{ }m/s$. The work done by the frictional force in time 2 s is $\_\_\_\_$ J.
(Take $g = 10\text{ }m/s^{2}$ )
Options
A.20
B.25
C.30
D.10
Solution
Sol. [Key Concept: Friction $f = mgsin30^{\circ}$; the (vertical) displacement is $8$ m, so $W = fdcos60^{\circ}$.]
$$f = mgsin30 = 5\text{ }N $$work done in 2 sec will be
$${d = 4 \times 2 = 8\text{ }m }{W = fdcos60^{\circ} }{W = (5)(8)cos60^{\circ} }$$$W = 20$ joule
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