Question

$\huge\mathfrak\red{Question:-}$


A block is projected upward on an inclined plane of inclination 37 along the line of greatest slope of u = .5 with the velocity of 5 m/ sec. The block 1 stop at the distance of from the starting point.​

Answer 1

$\purple{\bold{\underline{\underline{Answer:-}}}}$

$\green{\tt{\therefore{Distance\:travelled\:by\:block=1.25\:m}}}$

$\pink{\bold{\underline{\underline{Step-by-step\:explanation:-}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies {Initial \: velocity(u) = 5 \: m/s} \\ \\ \tt: \implies Angle \: of \: inclination = 37 \degree \\ \\ \tt: \implies \: Friction \: coefficient( \mu) = 0.5 \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: {\implies Distance \: travelled \: by \: block = ?}$

• According to the question :

$\text{Component \: of \: acceleration } \\ \tt: { \implies Perpendicular \: component \: to \: incline \: plane = g \: cos \theta} \\ \\ \tt: \implies Parallel \: component \: to \: incline \: plane = g \: sin \theta \\ \\ \tt\circ \: N = mg \: cos \: \theta \\ \\ \tt \circ \: fr = \mu N= mg \: cos \: \theta \\ \\ \tt \circ \: Decleration \: due \: to \: friction \: force = \mu g \: cos \: \theta \\ \\ \tt\because Total \: decleration = g \: sin \: \theta + \mu g \: cos \: \theta \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies {v}^{2} = {u}^{2} + 2as \\ \\ \tt\circ \:Final \: velocity = 0 \: m/s \\ \\ \tt: \implies {0}^{2} = {5}^{2} + 2 \times ( - g \: sin \: \theta - \mu g \: cos \: \theta) \times s \\ \\ \tt: \implies - 25 = - 2 \times (10 \times sin \:37 \degree + 0.5 \times 10 \times cos \: 37 \degree) \times s \\ \\ \tt: \implies 25 = 2 \times (10 \times \frac{3}{5} + 0.5 \times 10 \times \frac{4}{5} ) \times s \\ \\ \tt: \implies 25 = 2 \times (6 + 4) \times s \\ \\ \tt: \implies 25 = 20 \times s \\ \\ \tt: \implies s = \frac{25}{20} \\ \\ \green{\tt: \implies s = 1.25 \: m}\\\\ \green{\tt\therefore Distance\:travelled\:by\:block\:is\:\:1.25\:m}$

Answer 2

$\huge\underline{\bf \orange{AnSweR :}}$

• According to the question :

Component of acceleration

⟹ Perpendicular component to incline plane = gcosθ

⟹ Parallel component to incline plane = gsinθ

N = mgcosθ

fr = μN = mgcosθ

Decleration due to friction force = μgcosθ

∵Total decleration = gsinθ + μgcosθ

As we know that

⟹ v = u 2 +2as

Final velocity = 0m/s ⟹0 2 = 5 2 +2×(−gsinθ−μgcosθ)×s

⟹ −25=−2×(10×sin37°+0.5×10×cos37°)×s

⟹ 25=2×(10× 53 +0.5×10× 54 )×s

⟹ 25=2×(6+4)×s

=> 25=20×s

⟹ s = 2025

⟹ s = 1.25m

∴Distance travelled by block is1.25m