Question

On the top of cube of mass 'M1' and side 'a' another cube of mass 'M2' and side 'b' is lying. If the upper block is removed l, the decrease in potential energy of the system is (a) M2(a+b)g (b) M2 (a+b/2)g (c) M2(a/2+b)g (c) M2(a+b/2)g don't send the irrevelant answer otherwise it must be reported. okay​

Answer 1

Explanation:

Static Friction between the block is μ

k

M

1

g=.25×100=25N

static friction < applied force

so in this case kinetic friction will be act

f

k

=.12×100N=12N

FBD diagram of both the block as shown

for M

1

block which as acceleration a

1

F−f

K

=10a

1

eq(1)..

40−12=10a

1

a_{1}=2.8ms^{-2}

for M

2

slab which as acceleration

f

k

=30a

2

eq(2)..

12=30a

2

a

2

=0.4ms

−2

solving eq(1)and eq(2)

accleration of slab is 0.4ms

−2

Hence the B option is correct.

Answer 2

Answer:

nahi ans nahi dena to pls hat jao koi or de d ghar pls meri exam h