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

PhysicsW.E.P.Conservative force & potential energyEasy2 minPYQ_2020

PhysicsEasynumerical

A block of mass $m = 1 kg$ moving on a horizontal surface with speed $u = 2 m s^{- 1}$ enters a rough horizontal patch ranging from $x = 0.10 m$ to $x = 2.00 m$ . If the retarding force $f_{r}$ on the block in this range is inversely proportional to $x$ over this range i.e.
$f_{r} = - \frac{k}{x}$ for $0 .10 < x < 2 .00$
$f_{r} = 0$ for $x < 0.10$ and $x > 2.00$
If $k = 0.5 J$ then, the speed of this block ( in $m s^{- 1}$ ) as it crosses the patch is ( use $l n 20 = 3$ )

Answer:

1.00

Solution:

$W = \int f_{r} d x = - k \log_{e} \frac{2}{0.1} = - 1.5 J$

$∴ W = ∆ K$

$\frac{1}{2} \times 1 \times v^{2} - \frac{1}{2} \times 1 \times 4 = - 1.5$

$\Rightarrow v = 1 m s^{- 1}$

Stream:JEESubject:PhysicsTopic:W.E.P.Subtopic:Conservative force & potential energy

⏱ 2mℹ️ Source: PYQ_2020

Doubts & Discussion

Loading discussions...