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Two blocks with masses m1 = 10 kg and m2 = 5 kg are connected by a string that passes over a pulley, as shown in Figure 1. The coefficient of friction between the table and m2 is 0.01.

a) Calculate the acceleration of the system.
b) Determine the distance traveled by m2 in the first 5.0 seconds after the system is released.

Please provide the acceleration and distance traveled.

1 Answer

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Final answer:

The acceleration of the system is 3.90 m/s^2 and the distance traveled by m2 in the first 5.0 seconds after the system is released is 48.75 m.

Step-by-step explanation:

To calculate the acceleration of the system, we need to first consider the forces acting on the blocks. In this case, the force of gravity is acting on both blocks, and the tension in the string is also a force acting on the blocks. The equation for the net force on the system is:

net force = (m1 - m2) * g - friction force

where g is the acceleration due to gravity and the friction force is calculated as (coefficient of friction) * (normal force). Once we have the net force, we can use Newton's second law, F = ma, to solve for the acceleration. To determine the distance traveled by m2 in the first 5.0 seconds after the system is released, we can use the equation d = 1/2 * a * t^2, where d is the distance, a is the acceleration, and t is the time.

Given the information provided, we can now calculate the acceleration and distance traveled. The coefficient of friction is given as 0.01, so the friction force between the table and m2 is 0.01 * (m2 * g). Substituting in the given values, we get:
friction force = 0.01 * (5 kg * 9.8 m/s^2) = 0.49 N

Next, we can calculate the net force:
net force = (10 kg - 5 kg) * 9.8 m/s^2 - 0.49 N = 39.01 N

Finally, we can use Newton's second law to find the acceleration:
net force = m1 * a
39.01 N = 10 kg * a
a = 3.90 m/s^2

For part b, we can use the equation for distance traveled:
d = 1/2 * a * t^2
d = 1/2 * 3.90 m/s^2 * (5.0 s)^2
d = 48.75 m

User Steve Lianoglou
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