A 16.0 kg block is pulled at a constant speed on the table surface by a horizont

Question

A 16.0 kg block is pulled at a constant speed on the table surface by a horizontal cord with a tension of 30.0N. It was noticed that it initially required a tension of 35.0N just to get the block started moving.

a. draw a free body diagram to show this situation.
b. how large is the normal force of the table acting on the block?
c. how large is the kinetic friction force while the block slides on the table?
d. how large is the coefficient of kinetic friction?
e. if the block is pulled with a tension of 50.0N, how large will its acceleration be?

Explanation / Answer

a) diagram : block at center

normal force N Upwards

Weight W =m*g=downwards

FEXT=external force T  right wards

FFRICTIONAL FORCE=Ff left wards

(b)for this case of a horizontal surface ,the normal force 'N' turns out to be equal to the weight W because of sigma FY=W-N=m *ay=0

normal force =W=m*g=16*9.8=156.8 N

(c) kinetic frictional force=FFRIC=coefficient of kinetic friction*m*g

=coefficient *16*9.8=156.8*coefficient of kinetic friction

when T=30 N then coefficient of kinetic friction=

30=156.8*coefficient of kinetic friction

coefficient of kinetic friction=0.19132

when T=35N then coefficient of kinetic friction=0.2232

kinetic frictional force is=0.2232*35=34.99776 N

(D)coefficient of kinetic friction=0.2232

(e)to calculate acceleration:

50-(0.2232)*16*9.8=16*a

a=0.9376 m/sec^2

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