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Mila [183]
3 years ago
15

A block sliding along a horizontal frictionless surface with speed v collides with a spring and compresses it by 2.0 cm. What wi

ll be the compression if the same block collides with the spring at a speed of 2v?
Physics
1 answer:
Leviafan [203]3 years ago
7 0

Answer:

4.0 cm

Explanation:

For the compression of the spring, the kinetic energy of the mass equals the elastic potential energy of the spring.

So, 1/2mv² = 1/2kx² ⇒ x = (√m/k)v

Since m and k are constant since its the same spring x ∝ v

If our speed is now v₁ = 2v, our compression is x₁

x₁ = (√m/k)v₁ = (√m/k)2v = 2(√m/k)v = 2x

x₁ = 2x

Since x = 2.0 cm, our compression for speed = 2v is

x₁ = 2(2.0) = 4.0 cm

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Home Business Cottage and Small Industries Development Committee to organize National Industrial Goods and...

BUSINESS

Cottage and Small Industries Development Committee to organize National Industrial Goods and Technology Expo

By Glocal Khabar - 2368 0

National Industrial Goods and Technology Expo- Glocal Khabar

Kathmandu, February 19, 2018: Cottage and Small Industries Development Committee is set to organize the twenty-ninth edition of National Industrial Goods and Technology Expo from March 23, 2018 at Bhrikutimandap, Kathmandu.

The motto of the five-day event is ‘Let’s use home-made products, move ahead towards prosperity.’ The committee has shared that handicrafts, wool and bamboo products, goods made from handmade papers, different types of pickles and Palpali Dhaka would be put on display in the expo.

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3 years ago
If the velocity versus time graph of an object is a horizontal line, the object is
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Read 2 more answers
Use the values from PRACTICE IT to help you work this exercise. Suppose the same two vehicles are both traveling eastward, the c
Mariulka [41]

Answer:

A. v_{3}=12.17m/s

B. v_{car}=6.3m/s\\v_{truck}=-6.3m/s

C. ΔK=-4.13x10^3J

Explanation:

From the exercise we know that the car and the truck are traveling eastward. I'm going to name the car 1 and the truck 2

v_{1}=5.79m/s\\m_{1}=102kg\\v_{2}=18.5m/s\\m_{2}=103kg

A. Since the two vehicles become entangled the final mass is:

m_{3}=102kg+103kg=205kg

From linear momentum we got that:

p_{1}=p_{2}

m_{1}v_{1}+m_{2}v_{2}=m_{3}v_{3}

v_{3}=\frac{m_{1}v_{1}+m_{2}v_{2}}{m_{3} }=\frac{(102kg)(5.79m/s)+(103kg)(18.5m/s)}{(205kg)}

v_{3}=12.17m/s

B. The change in velocity of both vehicles are:

For the car

v_{car}=v_{f}-v_{o}=12.17m/s-5.79m/s=6.38m/s

For the truck

v_{truck}=12.17m/s-18.5m/s=-6.3m/s

C. The change in kinetic energy is:

ΔK=K_{2}-K_{1} =\frac{1}{2}m_{3}v_{3}^{2}-(\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2})

ΔK=\frac{1}{2}(205)(12.17)^{2}-(\frac{1}{2}(102)(5.79)^{2}+\frac{1}{2}(103)(18.5)^{2})=-4.13x10^{3}J

ΔK=-4.13x10^{3}J

6 0
3 years ago
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