Question

Work done by friction with respect to observer 1 in t = 2s. as shown in figure
Observer - 1
&
Observer - 2
10Kg
u=0.5
30Kg →F=400N
​

Answer 1

Answer:

very boring question.

Explanation:

not in the mood to answer. ask something interesting

Answer 2

Answer:

500J

Explanation:

a=ug

=0.5×10=5m/s^2

v/2=5

v=10

v^2-u^2=2as

100-0=10s

s=10

f=un

u×mg

0.5×10×10=50

work=f×s

50×10=500j

completed