Alarge box of mass m is pulled across a horizontal, frictionless surface by a horizontal rope with tension t. a small box of mass m sits on top of the large box. the coefficients of static and kinetic friction between the two boxes are μs and μk, respectively.
1- find an expression for the maximum tension t max for which the small box rides on top of the large box without slipping. express your answer in terms of the variablesm ,m , ms, and appropriate constants.2-a horizontal rope pulls a 10 kg wood sled across frictionless snow. a 6.0 kg wood box rides on the sled. what is the largest tension force for which the box doesn't slip? assume that mk= 0.50.

T = g μ_s ( M+m )

78.4 N

Explanation:

When both of them move with the same acceleration , small box will not slip over the bigger one. When we apply force on the lower box, it starts moving with respect to lower box. So a frictional force arises on the lower box which helps it too to go ahead . The maximum value that this force can attain is mg μ_s . As a reaction of this force, another force acts on the lower box in opposite direction .

Net force on the lower box

= T - mg μ_s = M a    ( a is the acceleration created by net force in M )

Considering force on the upper box

mg μ_s = ma

a = g μ_s

Put this value of a in the equation above

T - m gμ_s = M g μ_s

T = mg μ_s + M g μ_s

=  g μ_s ( M+m )

2 )

Largest tension required

T = 9.8 x  .50 x ( 10+6 )

= 78.4 N

the answer is 46 mls because when you add the speed per second it increses the speed

magnitude and direction