A block of metal of mass 2 kg on a horizontal table is attached to a mass of 0.4 kg by a light string passing over a frictionless pulley at the edge of the table. The block is subjected to a horizontal force by allowing the 0.45 kg mass to fall. The coefficient of sliding friction between the block and the table is 0.2. Calculatea. the initial accelerationb. the tension in the stringc. the distance the block would continue to move if after 2 s of motion, the string should break.
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Hii dear,
# Answers-
a) Initial acceleration = 0.2 m/s^2
b) Tension in the string = 4.32 N
c) Stopping distance = 0.0408 m
## Explaination-
# Given-
m1 = 0.45 kg
m2 = 2 kg
μ = 0.2
# Solution-
a) Initial acceleration-
a = (m1-μm2)g/(m1+m2)
a = (0.45-0.2×2)g/(0.45+2)
a = 0.2 m/s^2
b) Tension in the string-
T = μm2g + m2a
T = 0.2×2×9.8 + 2×0.2
T = 4.32 N
c) Stopping distance-
s = u^2/2μg ...u=at
s = (0.4^2)/(2×0.2×9.8)
s = 0.0408 m
s = 4.08 cm
Hope that helped you...
# Answers-
a) Initial acceleration = 0.2 m/s^2
b) Tension in the string = 4.32 N
c) Stopping distance = 0.0408 m
## Explaination-
# Given-
m1 = 0.45 kg
m2 = 2 kg
μ = 0.2
# Solution-
a) Initial acceleration-
a = (m1-μm2)g/(m1+m2)
a = (0.45-0.2×2)g/(0.45+2)
a = 0.2 m/s^2
b) Tension in the string-
T = μm2g + m2a
T = 0.2×2×9.8 + 2×0.2
T = 4.32 N
c) Stopping distance-
s = u^2/2μg ...u=at
s = (0.4^2)/(2×0.2×9.8)
s = 0.0408 m
s = 4.08 cm
Hope that helped you...
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