Question : A body of mass m is launched up on a rough inclined plane making an angle of 45° with horizontal. If the time of ascent is half of the time of descent, the fractional coefficient between plane and body is :
a) 2/5
b) 3/5
c) 3/4
d) 4/5
Doubt by Muskan
Ans : b) is the correct option.
Detailed Solution :
Case 1 : During Ascent
When the mass m is moving up the inclined plane
Acceleration of the body moving up the inclined plane is
a1=g(sinθ+µcosθ)
u1≠0
v1=0
s1=l
t1=?
We know,
s1=u1t+½a1t12
s1=u1t1-½a1t12 — (1)
Also
v1=u1+a1t1
0 = u1+(-a1)t1
-u1=-a1t1
u1=a1t1
substituting in equation (1)
s1=(a1t1)t1-½a1t12
s1=a1t12-½a1t12
s1=½a1t12 — (2)
Case 2 : During Descent
When the mass m is moving down the inclined plane
Acceleration of the body moving down the inclined plane is
a2=g(sinθ-µcosθ)
u2=0
v2≠0
s2=l
t2=?
s2=u2t2+½a2t22
s2=0×t2+½a2t22
s2=½a2t22 — (3)
t1=½t2 (Given) — (4)
equating equation (2) and (3)
½a1t12=½a2t22
a1t12=a2t22
g(sinθ+µcosθ)(½t2)2 = g(sinθ-µcosθ)(t2)2
[Using equation (4)]
(sinθ+µcosθ)¼t22 = (sinθ-µcosθ)t22
(sinθ+µcosθ)¼ = (sinθ-µcosθ)
(sinθ+µcosθ) = 4(sinθ-µcosθ)
sinθ+µcosθ = 4sinθ-4µcosθ
µcosθ+4µcosθ = 4sinθ-sinθ
5µcosθ = 3sinθ
5µ = 3(sinθ/cosθ)
5µ = 3tanθ
5µ = 3tan45°
5µ = 3(1)
µ = 3/5
Hence, b) would be the correct option.