How many times the following loop execute?
int x = 1;
while (x < 10)
{
if (x % 2 ==0)
System.out.println(x);
Answers
Answer:
Giventhatacceleration,a=
2
F
(M+m)
Nowfor(a):−
Weightofthesystem=(M+m)g=N
WhereNisthenormalforce.
TheforceofFriction=μNAlsogiven,
a=
2
F
(m+M)
Nowforequationofforce,alongitssurface
F−μN=(m+M)a
F−μ(m+M)g=(m+M)
2
F
(M+m)
F−μ(m+M)g=
m
F
2F−2μg(m+M)=F
μ=
2g(m+M)
F
HencethecoefficientofKineticfrcitionis
2g(m+M)
F
Nowfor(b):−
Frictional force:-Let the frictional force be f acting on the small block so the acceleration must be
a=
2
F
(m+M)
Sotheforce=m×a
f=m
2
M
(m+M)
f=m
2
M
(m+M)
Hencethefrictionalforceism
2
F
(m+M).
Nowfor(c):−Workdone:−Thevelocityoftheblockduringdisplacementd,
v²=u²+2ad
v²=2ad
sinceu=0
v²=2F
2
d
(m+M)
v²=Fd(m+M)
NowthefinalKineticenergy
K.E=
2
1
(mv²)
K.E=mF
2
d
(m+M)
Sotheworkdoneonsmallerblockislargerthanbiggerblockbythefrictionalforce,
sochangeinK.EismF
2
d
(m+M)