First slide

Friction on inclined plane of angle more than angle of repose

Question

A block slides down an inclined plane of slope of angle q with a constant velocity v' It is then projected up the plane with an initial velocity u. the distance upto which it will rise before coming to rest is

Moderate

A

$\frac{u^{2}}{4 gsin θ}$
B

$\frac{u}{4 gsin θ}$
C

$\frac{u^{2} \sin θ}{4 g}$
D

$\frac{usin θ}{4 g}$

Solution

The different forces in the two ca6es are shown in fig.
Case (i). As the block slides down with constant velocity, the acceleration is zero. In this case
$f = mgsin θ and f = μR = μmgcos θ$
$∴ μmgcos θ = mgsin θ$
$μ = \tan θ$ … (1)
Case (ii). The block is projected upward with initial velocity u and hence it experiences downward acceleration a. In this case
$mgsin θ + μmgcos θ = ma$
$or mgsin θ + mgtan θcos θ = ma$
$or mg (\sin θ + \sin θ) = ma$
$∴ a = 2 gsin θ . . . (2)$
Let x be the distance moved up the plane before the block comes to rest, Now
$\begin{aligned} v^{2} - u^{2} & = 2 as \\ 0 - u^{2} & = 2 (- 2 gsin θ) x \\ x & = \frac{u^{2}}{4 gsin θ} \end{aligned}$

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