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A rope is used to pull a 4.36 kg block at constant speed 4.33 m along a horizontal floor. The force on the block from the rope is 7.97 N and directed 26.2° above the horizontal. What are (a) the work done by the rope's force, (b) the increase in thermal energy of the block-floor system, and (c) the coefficient of kinetic friction between the block and floor?

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Answer:

Explanation:

The horizontal component of force applied = 7.97 cos 26.2 = 7.15 N.

Vertical component in upward direction = 7.97 sin 26.2 = 3.52 N.

Since the body is moving with uniform velocity, friction force will equalize the external horizontal component = 7.15 N.

So frictional force  = 7.15 N.

a) Work done by rope force per second = force in horizontal direction x displacement per second = 7.15 x 4.33 = 30.95 J

b) Increase in thermal energy per second will be due to negative work done by frictional force = work done by external force = 30.95 J.

c) Normal force acting downwards = weight - vertical component of external force  = 4.36 x 9.8 - 3.52 = 39.21 N

coefficient of friction = friction force / normal force = 7.15 / 39.21 = 0.18

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