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A body of mass 5 grams is launched up on a rough inclined plane making angle 30° with horizontal. The cofficient of friction between the body and the plane if the time of ascent is half the time of descent: a)3/5 tan60° b)3/5 tan30° c)4/5 tan60° d)4/5 tan30°

A body of mass 5 grams is launched up on a rough inclined plane making angle 30° with horizontal. The cofficient of friction between the body and the plane if the time of ascent is half the time of descent:
a)3/5 tan60°
b)3/5 tan30°
c)4/5 tan60°
d)4/5 tan30°

Grade:11

1 Answers

Arun
25750 Points
4 years ago

Sign convention-Down the slope(-) and up the slope(+)

-For upward motion,

A = Inclination = 30o.

So, weight component acting down the slope = F1 = -mgsinA

Normal reaction = mgcosA

Coefficient of friction = u

So, frictional force(Acting down the slope) = F2 = -umgcosA

So net force = F1 + F2 = -mg(sinA + ucosA)

Acceleration = -g(sinA + ucosA) = a1

As accn is uniform, eqn of motion are applicable.

Let s = displacement up the slope

Launch velocity = u; ta = time of ascent and td=time of descent

So, ta = 1/2td [Acc. to qts]----------(1)

So, s = uta+ 1/2a1ta2.

or s = uta + 1/2[-g(sinA + ucosA)]ta2.----------(2)

Now, final velocity = v =0

Also, v = u + a1ta

or 0 = u + [-g(sinA + ucosA)]ta.

or u = g(sinA + ucosA)ta.

Put u = g(sinA + ucosA)ta in (2) we get,

s = g(sinA + ucosA)ta2 - 1/2[g(sinA + ucosA)]ta2.

or s = 1/2[g(sinA + ucosA)]ta2

or s = 1/2[g(sinA + ucosA)](1/2td)2 [From(1)]

or s =1/8[g(sinA + ucosA)]td2.--------------(3)

For downward motion,

Weight component(acting down the slope) = -mgsinA = F1

frictional force(acting up the slope) = umgcosA = F2.

So, net force = F1 + F2 = mg(-sinA + ucosA)

Accn = a2 = g(ucosA - sinA)

As accn is uniform, eqn of motion are applicable.

Displacement(Down the slope) = -s

time of descent = td.

So, -s = utd + 1/2a2td2

As the downard motion starts with u = 0, the above expression becomes,

-s = 1/2a2td2= 1/2[g(ucosA - sinA)]td2------------(4)

Adding (3) and (4) we get,

1/8[g(sinA + ucosA)]td2 + 1/2[g(ucosA - sinA)]td2 = 0

or1/8[g(sinA + ucosA)]td2 = -1/2[g(ucosA - sinA)]td2.

or1/8[g(sinA + ucosA)]td2= 1/2[g(sinA - ucosA)]td2.

or 4sinA - 4ucosA = sinA + ucosA

or u = 3/5tanA = 3/5tan(30o) [As A = 30o]

or u = (3)1/2/5

Hence, the coefficient of friction = (3)1/2/5

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