Kinetic energy depends on the height of an object above the ground?

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

konstantin123 [22]

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

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

8 0

3 years ago

A 100 kg bag of sand has a weight on 100 N. When dropped its acceleration is what?

Anettt [7]

Answer:

The acceleration of the bag of sand is $1\ m/s^2$ .

Explanation:

We have,

Mass of a bag of sand is 100 kg

Weight of the bag of sand is 100 N

It is required to find the acceleration of the bag when it is dropped. The weight of an object is given by :

$F =ma$

When it is dropped, a = g

$g=\dfrac{F}{m}\\\\g=\dfrac{100}{100}\\\\g=1\ m/s^2$

So, the acceleration of the bag of sand is $1\ m/s^2$ .

4 0

4 years ago

A long electric cable is suspended above the earth and carries a current of 345 A parallel to the surface of the earth. The eart

jok3333 [9.3K]

Answer:

0.906 N

Explanation:

Formula for magnetic force acting on current carrying cable:

$F = IBLsin(\theta)$

Where I = 345A is the current in the wire, B = $5.6*10^{-5} T$ is the magnetic magnitude generated by Earth. L = 46.9 m is the cable length. $\theta = 88.2^o$ is the angle between vector B and cable direction.

$F = 345*5.6*10^{-5}*46.9*sin(88.2^o)$

$F = 0.906 N$

7 0

3 years ago

A spring is 20cm long is stretched to 25cm by a load of 50N. What will be its length when stretched by 100N. assuming that the e

IgorLugansk [536]

Answer:

Final Length = 30 cm

Explanation:

The relationship between the force applied on a string and its stretching length, within the elastic limit, is given by Hooke's Law:

F = kΔx

where,

F = Force applied

k = spring constant

Δx = change in length of spring

First, we find the spring constant of the spring. For this purpose, we have the following data:

F = 50 N

Δx = change in length = 25 cm - 20 cm = 5 cm = 0.05 m

Therefore,

50 N = k(0.05 m)

k = 50 N/0.05 m

k = 1000 N/m

Now, we find the change in its length for F = 100 N:

100 N = (1000 N/m)Δx

Δx = (100 N)/(1000 N/m)

Δx = 0.1 m = 10 cm

but,

Δx = Final Length - Initial Length

10 cm = Final Length - 20 cm

Final Length = 10 cm + 20 cm

<u>Final Length = 30 cm</u>

6 0

3 years ago

6. The hole on a level, elevated golf green is a horizontal distance of 150 m from the tee and at an elevation of 12.4 m above t

Georgia [21]

Answer:

u = 104.68 m/s

Explanation:

given,

horizontal distance = 150 m

elevation of 12.4 m

angle = 8.6°

horizontal motion = x = u cos θ. t .............(1)

vertical motion =

$y = u sin \theta - \dfrac{1}{2}gt^2$ ................(2)

from equation(1) and (2)

$y = x tan \theta - \dfrac{gx^2}{2u^2cos^2\theta}$ ..........{3}

$12.4 = 150\times tan (8.6) - \dfrac{9.8\times 150^2}{2u^2cos^2(8.6)}$

$\dfrac{9.8\times 150^2}{2u^2cos^2(8.6)} = 10.29$

$\dfrac{9.8\times 150^2}{2\times 10.29\times cos^2(8.6)} = u^2$

$u = \sqrt{10959.34}$

u = 104.68 m/s

The initial speed of the ball is u = 104.68 m/s

8 0

3 years ago