Question

A block of mass 5 kg resting on a horizontal surface is connected by a cord, passing over a light frictionless pulley to a hanging block of mass 5 kg. The coefficient of kinetic friction between the block and the surface is 0.5. Tension in the cord is : (g = 9.8 m/sec²)(a) 49 N (b) Zero(c) 36.75 N (d) 2.45 N

Answer 1

Answer:

c) 36.75 N

Explanation:

Mass of the block = 5kg

Coefficient of kinetic friction between the block and the surface = 0.5

Thus, as per the diagram -

Mg-T = Ma

5×9.8-T = 5×a  eq 1

For the block on the table -

T - Frictional force = Ma

T-uMg = Ma

T-0.5×5×9.8 = 5a  eq 2

Solving equation 1 and 2

a = 2.45ms-2

= 36.75N

Thus the tension in the cord will be 36.75

Answer 2

Answer:

for horizontal block

T - f = 5a => f=0.5×5×9.8= 24.5N

T-24.5 =5a.......(1)

for hanging block

5g-T=5a.......(2)

equating both 1 and 2 equations

T-24.5=49-T

2T=73.5

T=36.75N

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