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kvasek [131]
3 years ago
5

An apple is whirled round in a horizontal circle on the end of a string which is tied to the stalk. It is whirled faster and fas

ter and at a certain speed the apples is torn from the stalk.Why?
Physics
1 answer:
jek_recluse [69]3 years ago
6 0

when the apple moves in a horizontal circle, the tension force in the string provides the necessary centripetal force to move in circle. the tension in the string is given as

T=mv²/r

where T = tension force in the string , m = mass of the apple

v = speed of apple , r = radius of circle.

clearly , tension force depends on the square of the speed. hence greater the speed, greater will be the tension force.

at some point , the speed becomes large enough that it makes the tension force in the string becomes greater than the tensile strength of the string. at that point , the string breaks

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A 25 kg child plays on a swing having support ropes that are 2.20 m long. A friend pulls her back until the ropes are 42◦ from t
Semmy [17]

Answer:

A) P.E = 138.44 J

B) The velocity of swing at bottom, v = 3.33 m/s

C) The work done, W = -138.44 J

Explanation:

Given,

The mass of the child, m = 25 Kg

The length of the swing rope, L = 2.2 m

The angle of the swing to the vertical position, ∅ = 42°

A) The potential energy at the initial position ∅ = 42° is given by the relation

                                P.E = mgh joule

Considering h  = 0 for the vertical position

The h at ∅ = 42° is  h = L (1 - cos∅)

                               P.E = mgL (1 - cos∅)

Substituting the given values in the above equation

                               P.E = 25 x 9.8 x 2.2 (1 - cos42°)

                                      = 138.44 J

The potential energy for the child just as she is released, compared to the potential energy at the bottom of the swing is, P.E = 138.44 J

B) The velocity of the swing at the bottom.

At bottom of the swing the P.E is completely transformed into the K.E

                  ∴                 K.E = P.E

                                     1/2 mv² = 138.44

                                     1/2 x 25 x v² 138.44

                                            v² = 11.0752

                                             v = 3.33 m/s

The velocity of the swing at the bottom is, v = 3.33 m/s

C) The work done by the tension in the rope from initial position to the bottom

             Tension on string, T = Force acting on the swing, F

                      W=L\int\limits^0_\phi{F} \, d \phi

                             =L\int\limits^0_\phi{mg.sin \phi} \, d \phi

                            = -Lmg[cos\phi]_{42}^{0}

                            = - 2.2 x 25 x 9.8 [cos0 - cos 42°]

                            = - 138.44 J

The negative sign in the in energy is that the work done is towards the gravitational force of attraction.

The work done by the tension in the ropes as the child swings from the initial position to the bottom of the swing, W = - 138.44 J

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