A block of mass m. = 13 kg, on a frictionless inclined plane with an angle of 6

Question

A block of mass m. = 13 kg, on a frictionless inclined plane with an angle of 6 = 41° with the horizontal, is connected to another block of mass m2 = 4 kg by a massless string that passes over a pulley as shown in the figure. Take the pulley as an ideal pulley. Calculate the acceleration of the blocks (in m/s2) and the tension in the string (in N). (Enter the magnitudes. Due to the nature of this problem, do not use rounded intermediate values in your calculations—including answers submitted in WebAssign.) 31 e = 41° | m/s2 IIII

Explanation / Answer

Given,

m1 = 13 kg; m2 = 4 kg ; theta = 41 deg ;

for m2 we can write

m2g - T = m2a

for m1:

T - m1g sin(theta) = m1a

solving the abive two simultaneously for a we get

a = (m2g - m1 g sin(theta))/(m1 + m2)

a = (4 x 9.81 - 13 x 9.81 x sin41))/(13 + 4) = -2.611 m/s^2

Hence, a = 2.61 m/s^2

We have, m2g - T = m2a => T = m2 (g - a)

T = 4 (9.81 + 2.61) = 49.68 N

Hence, T = 49.68 N

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