Question

please don't spam kindly I beg you the one who knows answer kindly answer it​

Answer 1

Answer:

REF. Image

Let the displacement of block be

s. Then,

work force = −f

s

.s

consider the FBD

2Mg−T=2mg ____ (i)

T−μmg=ma

⇒T−

2

mg

=ma ___ (ii)

Adding (i) and (ii), we get

3ma=

2

3mg

a=g/2

To find displacement at v = 10 m/s

Applying 3

rd

eq

n

of motion,

v

2

−u

2

=2as

⇒(100)−0=2×

2

g

s

⇒s=

g

100

=

10

100

=10m

work-done by friction = (−

2

mg

)s

=−

2

10

×10=−50J

Answer 2

Answer:

sis ☺️

abhi good night nahi ho rahi hai good morning ho rahi hai