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Physics - W.E.P. Question with Solution | TestHub

PhysicsW.E.P.Calculation of workMedium2 minQB

PhysicsMediumsingle choice

Passage / Comprehension

A block of mass 10 kg is put gently on a belt-conveyor system of infinite length at t = 0 sec, which is moving with constant speed 20 m/sec rightward at all time, irrespective of any situation by means of a motor-system as shown in the figure. A constant force of magnitude 15 N is applied on the block continuously during its motion. ( $μ_{s} = 0.4, μ_{k} = 0.25$ )

Work done by the kinetic friction on the block of mass 10 kg is :-

Options:

Answer:

Solution:

While the block slips, kinetic friction $f_{k} = 0.25 \times 10 \times 10 = 25 N$ pushes it forward, together with the applied 15 N:

$a = \frac{25 + 15}{10} = 4 m/ s^{2}, t = \frac{20}{4} = 5 s, s = \frac{1}{2} \times 4 \times 5^{2} = 50 m$

$W_{f_{k}} = f_{k} s = 25 \times 50 = + 1250 J$

Ans. (A)

Stream:JEESubject:PhysicsTopic:W.E.P.Subtopic:Calculation of work

⏱ 2mℹ️ Source: QB

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