A 4.0 kg toboggan rests on a frictionless icy surface, anda 2.0 kg block rests o

Question

A 4.0 kg toboggan rests on a frictionless icy surface, anda 2.0 kg block rests on

top of the toboggan. The coefficient of static friction ms between the block and

the surface of the toboggan is 0.58, whereas the kinetic friction coefficient is

0.48. The block is pulled by a 30 N-horizontal force as shown.

a) Calculate the block's acceleration

b) calculate the toboggan's aceleration

c) if the applied force is gradually reduced, at what value do the block and the toboggan have the same acceleration.

A 4.0 kg toboggan rests on a frictionless icy surface, and a 2.0 kg block rests on top of the toboggan. The coefficient of static friction ms between the block and the surface of the toboggan is 0.58, whereas the kinetic friction coefficient is 0.48. The block is pulled by a 30 N-horizontal force as shown. Calculate the block's acceleration calculate the toboggan's acceleration if the applied force is gradually reduced, at what value do the block and the toboggan have the same acceleration.

Explanation / Answer

max static friction = ustatic mg = 0.58*2*9.81=11.38<30 so block will move

a) for ice

P - friction = ma

30 - 0.48*2*9.81 = 2 a

a=10.29

b) for tobooggan

friction = M a
0.48*2*9.81 = 4 a

a=2.35

c) now we don't know P but we knoe a ice = a toboggona

so a = 0.48*2*9.81/4=2.35

P = 2*2.35+0.48*2*9.81= 14.11 N

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