Physics, 25.01.2020 21:31 dustin112000

An object of mass 10 kg is released at point a, slides to the bottom of the 30° incline, then collides with a horizontal massless spring, compressing it a maximum distance of 0.75 m. the spring constant is 500 m/m, the height of the incline is 2.0 m, and the horizontal surface is frictionless. (a) what is the speed of the object at the bottom of the incline? (b) what is the work of friction on the object while it is on the incline? (c) the spring recoils and sends the object back toward the incline. what is the speed of the object when it reaches the base of the incline?

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Answer from: rndschopplein

Explanation:

The energy stored in the spring = the kinetic energy at the bottom of the incline

1/2 kx² = 1/2 mv²

kx² = mv²

(500 N/m) (0.75 m)² = (10 kg) v²

v ≈ 5.3 m/s

The energy stored in the spring = the initial potential energy - work done by friction

1/2 kx² = mgh - W

1/2 (500 N/m) (0.75 m)² = (10 kg) (9.8 m/s²) (2.0 m) - W

W ≈ 55 J

Since the horizontal surface is frictionless, the object will have the same speed at the bottom of the incline as before.

v ≈ 5.3 m/s

Answer from: samyajones68

In this problem we need to use the conservation of energy theorem, which is about to constant exchange between kinetic and potential energy. In this case, we have two important points to analyse. The first one is at the initial point, where the mass is just released at 2 m from the ground. The final point would be when the mass hits the massless spring.

So, in the initial point the total energy of the mass would be only potential energy due to its position and its lack of speed. Then, the mass will be released gaining speed, until reaches the spring, where all the kinetic energy is transformed in potential energy but due to elastic forces (spring). So, in the final point, both energy are the same, the kinetic and the elastic potential energy, which we express like this:

$K_{f}=U_{f}$

But, the definition of each one is:

$K=\frac{1}{2}mv^{2} \\U=\frac{1}{2}kx^{2}$

Where k is the constant of the spring. So, we replace this in the initial equation and solve for v:

$\frac{1}{2}mv_{f} ^{2}=\frac{1}{2}kx^{2}\\v_{f}=\sqrt{\frac{kx^{2}}{m}}\\v_{f}=\sqrt{\frac{500(0.75)^{2} }{10}}=5.30m/s$

Therefore, the speed in the spring point is 5.30 m/s.

Now, to calculate the work due to friction, we have to analyse first, because the problem doesn't offer a friction coefficient, but, we can analyse according to energies again, because work and energy are magnitudes close related, and they have the same dimension.

So, if we thought about it, we'll find that the work due to friction is what slows down the mass, subtracting its potential energy from the initial point, giving as a result the final potential energy:

$U_{f}=U_{i}-W_{k}$

$\frac{1}{2}kx^{2} =mgy-W_{k} \\W_{k}=10(9.8)(2)-\frac{1}{2}(500)(0.75)^{2}\\W_{k}=196-140.625=55.38J$

Therefore, the word due to friction is 55.38 J.

Lastly, the option c is asking about the speed of the object when it reaches the vase of the incline, which is the horizontal ground. The problem specify that this portion of the trajectory is frictionless, which mean that nothing slows down the mass.

Therefore, the speed in the horizontal part is the same 5.30 m/s.

Answer from: Quest

increases it

Answer from: Quest

e/c, where e is its total energy

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