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Larissa Fernandes

At what rate will a 7.5kg block accelerate if a 24 N force is applied to it, and the frictional force opposing it is 5.3N?

I’m trying to understand the acceleration of a 7.5kg block when a 24 N force is applied to it, and there’s a frictional force of 5.3N opposing it. Can you explain this to me?

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5 Answers

  1. The acceleration of an object can be calculated using Newton’s second law, F = ma, where F is the net force and m is the mass of the object. In this case, the net force is the applied force minus the frictional force: F_net = 24 N – 5.3 N = 18.7 N. Now, we can calculate the acceleration: a = F_net / m = 18.7 N / 7.5 kg = 2.49 m/s². So, the block will accelerate at a rate of 2.49 m/s².

  2. To find the acceleration, we need to subtract the frictional force from the applied force. So, F_net = 24 N – 5.3 N = 18.7 N. Now, we can use Newton’s second law, F = ma, to calculate the acceleration: a = F_net / m = 18.7 N / 7.5 kg = 2.49 m/s². Therefore, the block will accelerate at a rate of 2.49 m/s².

  3. When you apply a force of 24 N to a 7.5kg block and there’s a frictional force of 5.3 N opposing it, you can find the net force by subtracting the frictional force from the applied force: F_net = 24 N – 5.3 N = 18.7 N. Now, you can calculate the acceleration using Newton’s second law, F = ma: a = F_net / m = 18.7 N / 7.5 kg = 2.49 m/s². This is the acceleration of the block.

  4. The acceleration of the 7.5kg block can be determined by subtracting the frictional force from the applied force. So, F_net = 24 N – 5.3 N = 18.7 N. Now, using Newton’s second law, F = ma, we can calculate the acceleration: a = F_net / m = 18.7 N / 7.5 kg = 2.49 m/s². Thus, the block will accelerate at a rate of 2.49 m/s².

  5. In this scenario, the acceleration of the 7.5kg block can be found by subtracting the frictional force from the applied force. F_net = 24 N – 5.3 N = 18.7 N. Using Newton’s second law, F = ma, we can calculate the acceleration: a = F_net / m = 18.7 N / 7.5 kg = 2.49 m/s². So, the block will accelerate at a rate of 2.49 m/s².

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