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Lilit [14]
3 years ago
13

A block of mass m = 2.0 kg slides head on into a spring of spring constant k = 260 N/m. When the block stops, it has compressed

the spring by 14 cm. The coefficient of kinetic friction between block and floor is 0.42. While the block is in contact with the spring and being brought to rest, what are
(a) the work done by the spring force and
(b) the increase in thermal energy of the block-floor system?
(c) What is the block's speed just as the block reaches the spring? Please show all steps.
Physics
1 answer:
nasty-shy [4]3 years ago
8 0

Answer:

a) Ws = 2.548 J

b) Wf = 1.153 J

c) v = 1.923 m / s

Explanation:

a) The work done by the spring force  

Ws =  ½ * k * x²

Ws =  ½ * 260 N/m *0.14² m  

Ws =  2.548J

b) The increase in thermal energy can by find using  

Et = Wf

Wf = µ * m *g * x  

Wf = 0.42 * 2.0 kg *9.8 m/s² * 0.14m

Wf = 1.153 J

c) The speed just as the block reaches can by fin using

EK = Ws + Et

Ek = ( 2.548 + 1.153 ) J = 3.7 J

Ek = ½ * m * v²

v² = 2* Ek / m

v = √[2 * 3.7 J / 2.0 kg]

v = 1.923 m / s

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4 0
2 years ago
vA 61.2-kg circus performer is fired from a cannon that is elevated at an angle of 57.8 ° above the horizontal. The cannon uses
dsp73

Answer:

The effective spring constant of the firing mechanism is 1808N/m.

Explanation:

First, we can use kinematics to obtain the initial velocity of the performer. Since we know the angle at which he was launched, the horizontal distance and the time in which it's traveled, we can calculate the speed by:

v_0_x=\frac{x}{t}\\ \\v_0\cos\theta=\frac{x}{t}\\\\v_0=\frac{x}{t\cos\theta}

(This is correct because the horizontal motion has acceleration zero). Then:

v_0=\frac{20.8m}{(2.60s)\cos57.8\°}\\\\v_0=15.0m/s

Now, we can use energy to obtain the spring constant of the firing mechanism. By the conservation of mechanical energy, considering the instant in which the elastic band is at its maximum stretch as t=0, and the instant in which the performer flies free of the bands as final time, we have:

E_0=E_f\\\\U_e=K\\\\\frac{1}{2}kx^2=\frac{1}{2}mv^2\\\\\implies k=\frac{mv^2}{x^2}

Then, plugging in the given values, we obtain:

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Finally, the effective spring constant of the firing mechanism is 1808N/m.

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3 years ago
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