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Tems11 [23]
3 years ago
15

A 2.7-kg block is released from rest and allowed to slide down a frictionless surface and into a spring. The far end of the spri

ng is attached to a wall, as shown. The initial height of the block is 0.54 m above the lowest part of the slide and the spring constant is 453 N/m.
(a) What is the block's speed when it is at a height of 0.25 m above the base of the slide?

(b) How far is the spring compressed?

(c) The spring sends the block back to the left. How high does the block rise?
Physics
1 answer:
exis [7]3 years ago
5 0

a) The speed of the block at a height of 0.25 m is 2.38 m/s

b) The compression of the spring is 0.25 m

c) The final height of the block is 0.54 m

Explanation:

a)

We can solve the problem by using the law of conservation of energy. In fact, the total mechanical energy (sum of kinetic+gravitational potential energy) must be conserved in absence of friction. So we can write:

U_i +K_i = U_f + K_f

where

U_i is the initial potential energy, at the top

K_i is the initial kinetic energy, at the top

U_f is the final potential energy, at halfway

K_f is the final kinetic energy, at halfway

The equation can be rewritten as

mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2

where:

m = 2.7 kg is the mass of the block

g=9.8 m/s^2 is the acceleration of gravity

h_i = 0.54 is the initial height

u = 0 is the initial speed

h_f = 0.25 m is the final height of the block

v is the final speed when the block is at a height of 0.25 m

Solving for v,

v=\sqrt{u^2+2g(h_i-h_f)}=\sqrt{0+2(9.8)(0.54-0.25)}=2.38 m/s

b)

The total mechanical energy of the block can be calculated from the initial conditions, and it is

E=K_i + U_i = 0 + mgh_i = (2.7)(9.8)(0.54)=14.3 J

At the bottom of the ramp, the gravitational potential energy has become zero (because the final heigth is zero), and all the energy has been converted into kinetic energy. However, then the block compresses the spring, and the maximum compression of the spring occurs when the block stops: at that moment, all the energy of the block has been converted into elastic potential energy of the spring. So we can write

E=E_e = \frac{1}{2}kx^2

where

k = 453 N/m is the spring constant

x is the compression of the spring

And solving for x, we find

x=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(14.3)}{453}}=0.25 m

c)

If there is no friction acting on the block, we can apply again the law of conservation of energy. This time, the initial energy is the elastic potential energy stored in the spring:

E=E_e = 14.3 J

while the final energy is the energy at the point of maximum height, where all the energy has been converted into gravitational potetial energy:

E=U_f = mg h_f

where h_f is the maximum height reached. Solving for this quantity, we find

h_f = \frac{E}{mg}=\frac{14.3}{(2.7)(9.8)}=0.54 m

which is the initial height: this is correct, because the total mechanical energy is conserved, so the block must return to its initial position.

Learn more about kinetic and potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

brainly.com/question/6536722

#LearnwithBrainly

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(a) = +5.38m (b) = -5.38m (c) = 1.246m (d) = +0.3771m.

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Note: At maximum compression, the blocks would be at rest.

½Kx² = mg(h + x)

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X = + 0.3771m or -0.3589m

But we can't have a negative compression value,

X = + 0.3771m

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