A 2-kg block on a horizontal surface connected to a 1-kg hanging mass over a pulley with kinetic friction μ_k = 0.2 on the table. Using g = 9.8 m/s^2, what are the acceleration and the tension?

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Multiple Choice

A 2-kg block on a horizontal surface connected to a 1-kg hanging mass over a pulley with kinetic friction μ_k = 0.2 on the table. Using g = 9.8 m/s^2, what are the acceleration and the tension?

Explanation:
When two masses are connected over a pulley, they share the same magnitude of acceleration. The mass on the table is pulled by the rope with tension T, while kinetic friction opposes its motion. The hanging mass feels its weight pulling downward and the rope’s tension pulling upward. Set up the forces along the direction of motion for each mass and use a for the common acceleration. Friction force on the table mass: f_k = μ_k m g = 0.2 × 2 × 9.8 = 3.92 N. For the block on the table: T − f_k = m1 a → T − 3.92 = 2a. For the hanging mass: m2 g − T = m2 a → 9.8 − T = 1 a. From the second, T = 9.8 − a. Plug into the first: (9.8 − a) − 3.92 = 2a → 5.88 − a = 2a → a = 1.96 m/s^2. Then T = 9.8 − a = 9.8 − 1.96 = 7.84 N. So the acceleration is about 1.96 m/s^2, and the tension is about 7.84 N.

When two masses are connected over a pulley, they share the same magnitude of acceleration. The mass on the table is pulled by the rope with tension T, while kinetic friction opposes its motion. The hanging mass feels its weight pulling downward and the rope’s tension pulling upward. Set up the forces along the direction of motion for each mass and use a for the common acceleration.

Friction force on the table mass: f_k = μ_k m g = 0.2 × 2 × 9.8 = 3.92 N.

For the block on the table: T − f_k = m1 a → T − 3.92 = 2a.

For the hanging mass: m2 g − T = m2 a → 9.8 − T = 1 a.

From the second, T = 9.8 − a. Plug into the first: (9.8 − a) − 3.92 = 2a → 5.88 − a = 2a → a = 1.96 m/s^2.

Then T = 9.8 − a = 9.8 − 1.96 = 7.84 N.

So the acceleration is about 1.96 m/s^2, and the tension is about 7.84 N.

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