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gayaneshka [121]
3 years ago
6

A 0.250 kgkg toy is undergoing SHM on the end of a horizontal spring with force constant 300 N/mN/m. When the toy is 0.0120 mm f

rom its equilibrium position, it is observed to have a speed of 0.400 m/s.Find
(a) the total energy of the object at any point in its motion,
(b) the amplitude of the motion, and (c) the maximum speed attained by the object during its motion.
Physics
1 answer:
konstantin123 [22]3 years ago
8 0

Answer:

(a) The total energy of the object at any point in its motion is 0.0416 J

(b) The amplitude of the motion is 0.0167 m

(c) The maximum speed attained by the object during its motion is 0.577 m/s

Explanation:

Given;

mass of the toy, m = 0.25 kg

force constant of the spring, k = 300 N/m

displacement of the toy, x = 0.012 m

speed of the toy, v = 0.4 m/s

(a) The total energy of the object at any point in its motion

E = ¹/₂mv² + ¹/₂kx²

E = ¹/₂ (0.25)(0.4)² + ¹/₂ (300)(0.012)²

E = 0.0416 J

(b) the amplitude of the motion

E = ¹/₂KA²

A = \sqrt{\frac{2E}{K} } \\\\A = \sqrt{\frac{2*0.0416}{300} } \\\\A = 0.0167 \ m

(c) the maximum speed attained by the object during its motion

E = \frac{1}{2} mv_{max}^2\\\\v_{max} = \sqrt{\frac{2E}{m} } \\\\v_{max} = \sqrt{\frac{2*0.0416}{0.25} } \\\\v_{max} = 0.577 \ m/s

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lord [1]

Distance covered is given as follows

1). 7 km North

2). 5 km North

3). 1 km East

Now total distance covered will be given as

d = d_1 + d_2 + d_3

d = 7 km + 5 km + 1 km

d = 13 km

Now in order to find the displacement we will show all with their directions

\vec d_1 = 7 + 5 = 12 km towards North

\vec d_2 = 1 km towards East

So total displacement is

\vec d = \vec d_1 + \vec d_2

\vec d = 12 \hat j + 1 \hat i

so net displacement will be

d = \sqrt{12^2 + 1^2} = 12.04 km

so displacement is 12.04 km

6 0
3 years ago
Help me, Ive been stuck for 15 minutes
natita [175]
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5 0
2 years ago
How to derive the fourth equation of motion?
Leya [2.2K]

Answer:

To derive the fourth equation of motion, first we have to consider the equation for acceleration and then to rearrange it. or v2 = u2 + 2as and this equation of motion can be used to find the final velocity or the distance travelled if the other values are given.

Explanation:

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6 0
3 years ago
the winner of a 10km road race took half an hour to complete the race. Calculate the average speed. Give your answer in metres p
leonid [27]
Hello there!

For this:

1). Convert 10km to meters!
2). Convert the 30 minutes into seconds!

3). Use the following formula to solve for speed!

speed= distance/time

Note: The units should automatically work out to m/s. :)

My goal is to make sure you understand the problem, which is why I won't be giving you the answer. It'll be more work now, but less work in the future! :)

Hope this helped!

-------------


DISCLAIMER: I am not a professional tutor or have any professional background in your subject. Please do not copy my work down, as that will only make things harder for you in the long run. Take the time to really understand this, and it'll make future problems easier. I am human, and may make mistakes, despite my best efforts. Again, I possess no professional background in your subject, so anything you do with my help will be your responsibility. Thank you for reading this, and have a wonderful day/night!
4 0
3 years ago
A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th
shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

F = ma

a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2}

Now, we can calculate the final speed of the crate at the end of 10.0 m:

v_{f}^{2} = v_{0}^{2} + 2ad_{1}                  

v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s    

For the next 10.5 meters we have frictional force:

F - F_{\mu} = ma

F - \mu mg = ma

So, the acceleration is:

a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

v_{f}^{2} = v_{0}^{2} + 2ad_{2}  

v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s  

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.  

I hope it helps you!                              

7 0
3 years ago
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