A A 5.00 kg block is pulled along a horizontal frictionless floor by a cord that

Question

A

A 5.00 kg block is pulled along a horizontal frictionless floor by a cord that exerts a force F = 10.8 N at an angle ? = 25.0

A 5.00 kg block is pulled along a horizontal frictionless floor by a cord that exerts a force F = 10.8 N at an angle ? = 25.0 degree above the horizontal, as shown in Fig. 4-29. Figure 4-29 What is the magnitude of the acceleration of the block? m/s2 The force F is slowly increased. What is its value just before the block is lifted (completely) off the floor? N What is the magnitude of the acceleration of the block just before it is lifted (completely) off the floor? m/s2

Explanation / Answer

(a) What is the magnitude of the block's acceleration.

10.8cos(25) = 9.79 N (horizontal component of force)

9.79/5 = 1.958 m/s^2

(b) The force magnitude F is slowly increased. What is its value just before the block is lifted (completely) off the floor.

This time find the vertical component:

mg = 5 x 9.8 = 49N (vertical component must equal mg)

49/sin(25) = 115.94 N

(c) What is the magnitude of the acceleration of the block just before it is lifted (completely) off the floor.

As question (a)

115.94cos(25) = horizontal force = 105 N

105/5 = 21 m/s^2

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