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 frictionless horizontal surface. The vertically hanging mass is 6.00 kg and the mass on the horizontal surface is 4.00 kg. The acceleration of the 4.00 kg mass is A) 1.98 m/s² to the left. B) 3.92 m/s² to the left. C) 5.88 m/s² to the left. D) 6.75 m/s² to the left. E) 7.84 m/s² to the left.

Question

Quizplus · Accepted Answer

We can use Newton's second law and free body diagrams to solve this problem. Let's draw a diagram for the two masses:

For the hanging mass, we have:

$$F_{T}-F_g=m_{1}a$$

where $F_T$ is the tension in the string and $F_g$ is the weight of the hanging mass. We can rearrange this and substitute in known values:

$$F_T=m_{1}g-m_{1}a=(6.00\ kg)(9.81\ m/s^2)-(6.00\ kg)a$$

For the sliding mass, we have:

$$F_{T}=m_{2}a$$

where $F_T$ is the tension in the string and $m_2$ is the mass of the sliding mass. We can substitute in known values:

$$F_T=(4.00\ kg)(11.0\ m/s^2)$$

Since the tension in the string is the same for both masses, we can set the two expressions for $F_T$ equal to each other:

$$(6.00\ kg)(9.81\ m/s^2)-(6.00\ kg)a=(4.00\ kg)(11.0\ m/s^2)$$

Solving for $a$, we get:

$$a=5.88\ m/s^2\ to\ the\ left$$

Therefore, the answer is (C) 5.88 m/s^2 to the left.

Two Masses Are Connected by a String Which Passes Over