Two masses are connected by a string which passes over a pulley with negligible mass and friction.One mass hangs vertically and one mass slides on a 30.0 degrees incline.The inclined surface has a coefficient of kinetic friction of 0.200.The vertically hanging mass is 4.00 kg and the mass on the incline is 6.00 kg.The acceleration of the 4.00 kg mass is (the initial velocity of the 6.00 kg mass is up the incline) A) 7.16 m/s². B) 5.84 m/s². C) 2.98 m/s². D) 1.53 m/s². E) 3.90 m/s².

Question

Quizplus · Accepted Answer

To solve the problem, we can use the following steps:

1. Draw a free body diagram for each mass.

2. Write the equations of motion for each mass using Newton's Second Law:

For the 4.00 kg mass:
m1*a = T - m1*g
where m1 is the mass of the 4.00 kg mass, a is its acceleration, T is the tension in the string and g is the acceleration due to gravity.

For the 6.00 kg mass:
m2*a = m2*g*sin(theta) - T - m2*g*cos(theta)*mu
where m2 is the mass of the 6.00 kg mass, a is its acceleration, T is the tension in the string, g is the acceleration due to gravity, theta is the angle of the incline and mu is the coefficient of kinetic friction.

3. Solve for the tension T by eliminating it from the two equations:

T = m2*g*sin(theta) - m2*g*cos(theta)*mu - m1*a

4. Substitute the expression for T into the equation for the 4.00 kg mass to obtain an expression for its acceleration:

m1*a = (m2*g*sin(theta) - m2*g*cos(theta)*mu - m1*a) - m1*g
(m1 + m2)*a = m2*g*(sin(theta) - mu*cos(theta)) - m1*g
a = (m2*g*(sin(theta) - mu*cos(theta)) - m1*g)/(m1 + m2)

5. Substitute the given values and solve for a:

a = (6.00 kg*9.81 m/s^2*(sin(30.0 degrees) - 0.200*cos(30.0 degrees)) - 4.00 kg*9.81 m/s^2)/(4.00 kg + 6.00 kg) = 5.84 m/s^2

Therefore, the acceleration of the 4.00 kg mass is 5.84 m/s^2, which corresponds to choice (B).

Two Masses Are Connected by a String Which Passes Over