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A strongman drags a 122kg tire across the ground at 1.20m/s^2. If there is 520N of friction between the tire and the ground, how much force does the strongman pull with?

User Tzerb
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Final answer:

The strongman must pull with a force of 666.4 newtons to drag the 122kg tire across the ground at an acceleration of 1.20m/s^2, considering there is 520N of friction to overcome.

Step-by-step explanation:

To calculate the force the strongman pulls with, we apply Newton's second law, which states that the force is the product of mass and acceleration (F = m × a). Given the mass (m) of the tire is 122kg and the acceleration (a) is 1.20m/s2, the force due to acceleration can be calculated as:

F = 122kg × 1.20m/s2 = 146.4N (force due to acceleration).

However, we must also overcome the force of friction, which is given as 520N. The total force the strongman needs to apply is the sum of the force due to acceleration and the frictional force:

Total force = Force due to acceleration + Frictional force

Total force = 146.4N + 520N = 666.4N.

Thus, the strongman must pull with a force of 666.4 newtons to drag the tire across the ground at the given acceleration of 1.20m/s2, overcoming the friction.

User Will Prescott
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