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andreev551 [17]
3 years ago
14

A block of 1 kg with a speed 1 m/s hits a spring placed horizontally as shown in the figure. If spring constant is 1000 N/m, fin

d the compression in the spring.
Physics
1 answer:
Evgen [1.6K]3 years ago
8 0

Answer:

0.0316 m

Explanation:

Wok done = Energy change

Work done on the spring = Energy change of the block

  Elastic Potential stored = Kinetic energy of the block

\frac{1}{2} kx^{2} = \frac{1}{2} mv^{2}

                              x = \sqrt{v^{2}\frac{m}{k}  }

                              x = \sqrt{1^{2}\frac{1}{1000}  }

                              x = 0.0316 m

k = spring constant

m = mass of block

v = velocity of the block

x = compression of spring

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The tungsten filament of a light bulb has an operating temperature of about 2 100 K. If the emitting area of the filament is 1.0
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75 W

Explanation:

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Using Stefan's law, Power output of the light bulb is given as

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A single Oreo cookie provides 53 kcal of energy. An athlete does an exercise that involves repeatedly lifting (without accelerat
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Answer:

Approximately 325 (rounded down,) assuming that g = 9.81\; {\rm N \cdot kg^{-1}}.

The number of repetitions would increase if efficiency increases.

Explanation:

Ensure that all quantities involved are in standard units:

Energy from the cookie (should be in joules, {\rm J}):

\begin{aligned} & 53\; {\rm kCal} \times \frac{1\; {\rm kJ}}{4.184\; {\rm kCal}} \times \frac{1000\; {\rm J}}{1\; {\rm kJ}} \approx 2.551 \times 10^{5}\; {\rm J} \end{aligned}.

Height of the weight (should be in meters, {\rm m}):

\begin{aligned} h &= 2\; {\rm dm} \times \frac{1\; {\rm m}}{10\; {\rm dm}} = 0.2\; {\rm m}\end{aligned}.

Energy required to lift the weight by \Delta h = 0.2\; {\rm m} without acceleration:

\begin{aligned} W &= m\, g\, \Delta h \\ &= 100\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times 0.2\; {\rm m} \\ &= 196\; {\rm N \cdot m} \\ &= 196\; {\rm J} \end{aligned}.

At an efficiency of 0.25, the actual amount of energy required to raise this weight to that height would be:

\begin{aligned} \text{Energy Input} &= \frac{\text{Useful Work Output}}{\text{Efficiency}} \\ &= \frac{196\; {\rm J}}{0.25} \\ &=784\; {\rm J}\end{aligned}.

Divide 2.551 \times 10^{5}\; {\rm J} by 784\; {\rm J} to find the number of times this weight could be lifted up within that energy budget:

\begin{aligned} \frac{2.551 \times 10^{5}\; {\rm J}}{784\; {\rm J}} &\approx 325 \end{aligned}.

Increasing the efficiency (the denominator) would reduce the amount of energy input required to achieve the same amount of useful work. Thus, the same energy budget would allow this weight to be lifted up for more times.

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2 years ago
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