A block slides down an inclined plane of slope θ with constant velocity. It is then projected up the plane with an initial speed v0. How far up the incline will it move before coming to rest?

Question

A block slides down an inclined plane of slope θ with constant velocity. It is then projected up the plane with an initial speed v0. How far up the incline will it move before coming to rest? (A) (v02/4 g sin θ) (B) (v02/g sin θ) (C) (v02/2 g sin θ) (D) (v02/2 g)

Tardigrade Admin · Accepted Answer

g sin θ=μ g cos θ ⇒ μ= tan θ When block is project up a= sin θ+μ g cos θ=2 g sin θ d stop =( v 02/2 a )=( v 02/2 × 2 g sin θ) =( V 02/4 g sin θ)

Q. A block slides down an inclined plane of slope $θ$ with constant velocity. It is then projected up the plane with an initial speed $v_{0}$ . How far up the incline will it move before coming to rest?

$\frac{v _{0}^{2}}{4 g s i n θ}$

$\frac{v _{0}^{2}}{g s i n θ}$

$\frac{v _{0}^{2}}{2 g s i n θ}$

$\frac{v _{0}^{2}}{2 g}$

$g sin θ = μg cos θ$
$\Rightarrow μ = tan θ$
When block is project up
$a = sin θ + μg cos θ = 2 g sin θ$
$d_{s t o p} = \frac{v _{0}^{2}}{2 a} = \frac{v _{0}^{2}}{2 \times 2 g s i n θ}$
$= \frac{V _{0}^{2}}{4 g s i n θ}$

Q. A block slides down an inclined plane of slope θ with constant velocity. It is then projected up the plane with an initial speed v0​. How far up the incline will it move before coming to rest?

4gsinθv02​​

gsinθv02​​

2gsinθv02​​

2gv02​​