A 5 kg block is pulled along a horizontal surface by a 28N force to the right. Compute the acceleration of the block if the surface is
a)
smooth,

b)
rough with a coefficient of kinetic friction of 0.23 between the block and the
surface.
[ 5.6ms 2,3.4ms]

(a) Without friction, Newton's second law tells us the net horizontal force is such that

28 N = (5 kg) a

a = (28 N) / (5 kg)

a = 5.6 m/s²

(b) With friction (magnitude f ), we get a net horizontal force such that

28 N - f = (5 kg) a

and a net vertical force such that

n - w = 0

since the block is only being pulled horizontally, where w is the weight of the block and n is the magnitude of the normal force. So

n = w = (5 kg) (9.8 m/s²)

n = 49 N

Friction is proportional to the normal force such that

f = µ n

We're given µ = 0.23, so

f = 0.23 (49 N)

f ≈ 11 N

Solve for a :

28 N - 11 N = (5 kg) a

a = (17 N) / (5 kg)

a = 3.4 m/s²

pe = weightxheight = 2000x1000=2,000,000joules

part 1: which substance has a higher solubility?

a

smallest amount left undissolved atall temps in table shown most has dissolved high sol'by