Two blocks with masses m_1 = 1.20 kg and m_2 = 3.30 kg are connected by a massle

Question

Two blocks with masses m_1 = 1.20 kg and m_2 = 3.30 kg are connected by a massless string, as shown in the Figure. They are released from rest. The coefficient of kinetic friction between the upper block and the surface is 0.230. Assume that the pulley has a negligible mass and is frictionless, and calculate the speed of the blocks after they have moved a distance 61.0 cm. Use Newton's second law to find the net force acting on both blocks. Since they are connected by a string, they act as one body and have the same acceleration. Review motion in one dimension to determine their speed.

Explanation / Answer

Let the acceleration of the system be a
Let the tension in the string be T

m2*g - T =m2*a
3.30*9.8 - T = 3.30*a ------1

T - uk*m1*g = m1*a
T - 0.230 * 1.20 * 9.8 = 1.20 * a ------2

Solving these eq
a =  6.59 m/s^2


Now speed after distance 0.61m,
Vf^2 = u^2 + 2*a*s
Vf = sqrt(2*6.59*0.61)
Vf = 2.84 m/s

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