Question

A 100 kg mass is pulled along a frictionless surface by a horizontal force F such that its acceleration is 10.0 m/s2. A 20 kg mass slides along the top of the 100 kg mass and has an acceleration of 3.0 m/s2. (It thus slides backward relative to the 100 kg mass.)

Answer 1

given,

mass = 100 kg

acceleration = 10 m/s²

A mass 20 kg slides over 100 kg block

acceleration = 3 m/s²

horizontal friction exerted by the 100 kg block on 20 kg

using newton's second law

F - f = 0

F = f

f = ma

f = 20 × 3

f = 60 N

now net force acting on the 100 kg block

F_net = m a

F_net = 100 x 10

F_net = 1000 N

after 20 kg block falls the acceleration of the bock

F = 1000 +60

F = 1060 N

acceleartion on the block

`a = (F)/(m)`

`a = (1060)/(100)`

a = 10.60 m/s²