physics help Physics help A 11 kg block is placed on top of a 34 kg block (see f

Question

physics help Physics help A 11 kg block is placed on top of a 34 kg block (see figure below). A force of 360 N is applied to the right on the lower block, and the upper block slips on the lower block (accelerating less than the lower block). The coefficient of kinetic friction between the upper block and the lower block is 0.2, and the coefficient of kinetic friction between the lower block and the floor is 0.46. (Assume the positive direction is to the right.) What is the acceleration of the upper block? (Indicate the direction with the sign of your answer.) m/s2 What is the acceleration of the lower block? (Indicate the direction with the sign of your answer.) m/s2 How big would the coefficient of static friction between the upper and lower block have to be so that the upper block would not slip on the lower block?

Explanation / Answer

a) F = ma ( second law of newton)

upper block only have friction acting on it

on applying this for m2

f1 = m2a ( f1 is kinetic friction b/w upper and lower block = uN)

.2*11*9.8 = 11a

=> a = 1.96 m/s2

b)

F = ma ( second law of newton)

on applying this for m1

F-f1-f2 = m1a ( f2 is kinetic friction b/w ground and lower block )

363 - .2*11*9.8- .47*45*9.8 = 34b

=> b = 3.94 m/s2

c) in this case both will move with same acc.

F-u(m1+m2)g = (m1 + m2)c

=> 363- .44*45*9.8 = 45c

=> c=   3.75 m/s^2

upper block only have static friction acting on it

now again use F= ma for upper block

we get f(static)= m2 c

um2g = m2c

=> u = 0.38

for this value of static friction both will move with same acc and no slipping will occure blw both block

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