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SVEN [57.7K]
3 years ago
5

A box of bananas weighing 40.0 N rests on a horizontal surface. The coefficient of static friction between the box and the surfa

ce is 0.40 and the coefficient of kinetic friction is 0.20.
(a) If no horizontal force is applied to the box and the box is at rest, how large is the friction force exerted on it?
(b) What is the magnitude of the friction force if a monkey applies a horizontal force of 6.0N to the box and the box is initially at rest?
Physics
1 answer:
Burka [1]3 years ago
8 0

Answer:

(a) = 0 N

(b) = 2.4 N

Explanation:

given

box of banana weight = 40.0 N

coefficient of static friction  μ = 0.40

coefficient of kinetic friction = 0.20

a).  when no horizontal force is applied .

according to Newton 's third law of motion If there is no force applied to the box,so the frictional  force exerted is 0 N

b) magnitude of friction force

box and the box is initially at rest

    friction force =.Static friction coefficient × weight of the box

      friction force = 0.40 × 6

       friction force =  2.4 N

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A ball of mass 0.120 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.820 m. W
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Answer:

1.0752 kgm/s

Explanation:

Considering when the drop was dropped from rest from a height,

mass of the ball, m = 0.120 kg

height, h = - 1.25 m

the initial velocity, u = 0 m/s

the acceleration due to gravity, g = - 9.8 m/s²

From equation of motion

                            V^{2} = U^{2} + 2gh

Substituting the values,

                             V^{2} = 0^{2} + 2(-9.8 m/s^{2})(-1.25 m)

                             V^{2} = 24.5 m/s

                             V = \sqrt{24.5} \ m/s

                             V = 4.95 \ m/s

                            V = ± 4.95 m/s

                            V = - 4.95 m/s

Since the ball is moving downward, the final velocity of the ball when it hits the floor is  V = - 4.95 m/s  

Considering when the ball rebounds from the floor,

assume the mass of the ball still remain, m = 0.120 kg

height, h = 0.820 m

the final velocity, v = 0 m/s  

the acceleration due to gravity, g = - 9.8 m/s²

From equation of motion

                            V^{2} = U^{2} + 2gh

Substituting the values,

                            0^{2} = U^{2} + 2(-9.8 m/s^{2})(0.820 m)

                            0 = U^{2} - 16.072 m/s

                            U^{2} = 16.072 m/s

                            U = \sqrt{16.072} \ m/s

                           U = ± 4.01 m/s

                          U = + 4.01 m/s

Since the ball is moving upward, the initial velocity of the ball from the bounce from the floor is  U = + 4.01 m/s                        

From Newton's second law of motion, applied force is directly proportional to the rate of change in momentum.

                            F = \frac{mv - mu}{t}

                          F.t = m(v - u)

       ⇒      Impulse = Change in momentum

To calculate the impulse, the moment before the ball hits the ground will be the initial momentum while the moment the ball rebounces will be the final velocity,                        

          ∴          F.t = 0.120  kg(4.01  m/s - (-4.95  m/s) )

                      F.t = 0.120  kg(4.01  m/s + 4.95  m/s) )

                      F.t = 0.120  kg × 8.96  m/s

                      Impulse  = 1.0752 kgm/s

The impulse given to the ball by the floor is 1.0752 kgm/s

                             

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Explanation:

Given

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Extension in a wire is given by

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where, E=\text{Youngs modulus}

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\Rightarrow \delta_2=\dfrac{FL}{A_2E}\quad \ldots(ii)

Divide (i) and (ii)

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german

The work done to stretch the spring will be 112 J.

<h3>What is spring force?</h3>

The force required to extend or compress a spring by some distance scales linearly with respect to that distance is known as the spring force. Its formula is

F = kx

The given data in the problem is;

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