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

$\blue{\bold{\underline{\underline{Answer:}}}}$

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

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

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

• According to given 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

Answer:

1.25m

Explanation:

Given,

Friction coefficient = 0.5

Friction coefficient = 0.5Angle of Inclination = 37°

In the block diagram of the motion for the block,

We observe that,

There are two forces acting download on block for against its motion due to force that is Friction force and it's mass in the direction of motion.

Now, if we consider that distance traveled by block in upward direction is s, then it's velocity at this point will be zero.

If the acceleration of the particle is 'a' then,

For equilibrium condition,

ma = Friction Force + Force due to mass of block

ma = μN + Mgsin37°

ma = μMgcos37° + Mgsin37°

a = μgcos37° + gsin37°

a = (0.5×10×0.7986) + (10×0.60)

a = 10 m/s²

Now, by using third equation of motion, v² = u² - 2as

0 = 5² - 2×10s

20s = 25

s = 25/20

s = 5/4

s = 1.25 m