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kotykmax [81]
3 years ago
14

To hoist himself into a tree, a 72.0-kg man ties one end of a nylon rope around his waist and throws the other end over a branch

of the tree. He then pulls downward on the free end of the rope with a force of 360 N. Neglect any friction between the rope and the branch, and determine the man’s upward acceleration. Use g =9.80 m/sec2.
Physics
1 answer:
sergij07 [2.7K]3 years ago
4 0

Answer:

man upward acceleration is 0.14m/s^2

Explanation:

given data:

mass of man = 72 kg

downward force = 360 N

The mass of man of weight 72 kg is hang from two sections of rope, one section pf rope ties around man waist and other section is ties in man hands. when he pulls down the rope  with 360 N force then each section of  rope pulls with 360 N

we know that

Weight= mass × gravity= 72kg × 9.8 = 705.6N

Force = mass× acceleration

Force= -705.6 + (2 × 358) = 10.4 N

acceleration = \frac{10.4}{72} = 0.14m/s^2

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A 0.37-kg object connected to a light spring with a force constant of 23.2 N/m oscillates on a frictionless horizontal surface.
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Answer:

a) v = 31.67 cm / s , b)   v = -29.36 cm / s , c) v= 29.36 cm/s, d) x = 3.46 cm

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         Ф = 0

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        x = 4.0 cos (7,918 t)

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Let's look for the time for a displacement of x = 1.5 cm, remember that the angles must be in radians

          7,918 t = cos⁻¹ 1.5 / 4.0

          t = 1,186 / 7,918

          t = 0.1498 s

We look for speed

         v = -31.67 sin (7,918 0.1498)

         v = -29.36 cm / s

c) if the object passes the equilibrium equilibrium position again at this point the velocity has the same module, but the opposite sign

         v = 29.36 cm / s

d) let's look for the time for the condition v = v_max / 2

         31.67 / 2 = 31.67 sin ( 7,918 t)

          7.918t = sin⁻¹ 0.5

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With this time let's look for displacement

         x = 4.0 cos (7,918 0.06613)

        x = 3.46 cm

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