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kupik [55]
4 years ago
12

A spring gun is made by compressing a spring in a tube and then latching the spring at the compressed position. A 4.87-g pellet

is placed against the compressed and latched spring. The spring latches at a compression of [04]____________________ cm, and it takes a force of 9.13 N to compress the spring to that point. How much work (J) is done by the spring when the latch is released and the pellet leaves the tube?

Physics
1 answer:
Serga [27]4 years ago
8 0

Complete Question

The complete question is shown on the first uploaded image

Answer:

The work done by the spring is = 0.27 J

Explanation:

Force = torque × length

Given

F = 9.13 N

length (L) = 5.91 cm  = 0.0591 m                         [Note 1 m = 100 cm ]

considering the formula above

          9.13 = k * 0.0591    where k denotes torque

          k = 154.48\ N/m

Energy Stored  = \frac{1}{2} k x^2

                          = \frac{1}{2} * 154.48 * (0.0591)^2

                         = 0.27J  

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Answer:

The answer is in the explanation

Explanation:

A)

i) The blocks will come to rest when all their initial kinetic energy is dissipated by the friction force acting on them. Since block A has higher initial kinetic energy, on account of having larger mass, therefore one can argue that block A will go farther befoe coming to rest.

ii) The force on friction acting on the blocks is proportional to their mass, since mass of block B is less than block A, the force of friction acting on block B is also less. Hence, one might argue that block B will go farther along the table before coming to rest.

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m_{A}\frac{\mathrm{d} v}{\mathrm{d} t} = -m_{A}g\nu_{s}\Rightarrow \frac{\mathrm{d} v}{\mathrm{d} t} = -\nu_{s}g \quad (1)

Here, \nu_{s} is the coefficient of friction between the block and the surface of the table. Equation (1) can be easily integrated to get

v(t) = C-\nu_{s}gt \quad (2)

Here, C is the constant of integration, which can be determined by using the initial condition

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\frac{\mathrm{d} x}{\mathrm{d} t} = v_{0}-\nu_{s}gt\Rightarrow x(t) = v_{0}t-\nu_{s}g\frac{t^{2}}{2}+D \quad (6)

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The distance travelled by block A before stopping is

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C) We can see that the expression for the distance travelled for block A is independent of its mass, therefore if we do the calculation for block B we will get the same result. Hence the reasoning for Student A and Student B are both correct, the effect of having larger initial energy due to larger mass is cancelled out by the effect of larger frictional force due to larger mass.

D)

i) The block A is moving in a circle of radius L+\frac{d}{2} , centered at the pivot, this is the distance of pivot from the center of mass of the block (assuming the block has uniform mass density). Because of circular motion there must be a centripetal force acting on the block in the radial direction, that must be provided by the tension in the string. Hence

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ii) The forces acting on the block are

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b) Friction: Acting tangentially, in the direction opposite to the velocity of the block at any given time, therefore it decreases the speed of the block.

The speed decreases linearly with time in the same manner as derived in part (C), using the expression for tension in part (D)(i) we can see that the tension in the string also decreases with time (in a quadratic manner to be specific).

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