All categories

2 years ago

A spring toy jumps up from the floor and comes back down to the floor. Its initial speed is 12 m/s. What is the

Physics

1 answer:

dybincka [34]2 years ago

8 0

Answer:

$0\; \rm m \cdot s^{-1}$ .

Explanation:

In a projectile motion, the kinetic energy (KE) of the projectile is converted to gravitational potential energy (GPE) and then from GPE back to KE.

In this example, the spring toy is the projectile. Refer to the diagram attached.

The spring toy started with a velocity of $12\; \rm m\cdot s^{-1}$ , meaning that its initial KE is non-zero.
On the way up, the KE of this spring toy gets converted to GPE.
At the top of the trajectory, the GPE of this toy is maximized while its KE is minimized (zero).
As the toy returns to the ground, the GPE of this toy gets converted back to KE.
The GPE of this toy is zero when the toy is on the ground.

In other words, the kinetic energy (KE) of this toy would be $0$ when it is at the top of the trajectory and GPE is maximized.

Since the KE of this toy at the top of the trajectory would be $0$ , the velocity of this toy at that moment would also be $0\!$ .

You might be interested in

A ball rolls over the edge of a platform with only a horizontal velocity. The height of the platform is 1.60m and the horizontal

AysviL [449]

Answer:

v = 46.99 m/s

Explanation:

The velocity of the ball just before it touches the ground, is given by the following formula:

$v=\sqrt{v_x^2+v_y^2}$ (1)

vx: horizontal component of the velocity

vy: vertical component of the velocity

The vertical component vy is calculated by using the following formula:

$v_y^2=v_{oy}^2+2gh$ (2)

vy: final velocity

voy: initial vertilal velocity = 0m/s (because it is a semi parabolic motion)

g: gravitational acceleration = 9.8 m/s^2

h: height = 1.60m

You replace the values of the parameters in the equation (2):

$v_y=2(9.8m/s^2)(1.60m)=31.36\frac{m}{s}$

vx is calculated by using the information about the horizontal range of the ball:

$R=v_o\sqrt{\frac{2h}{g}}$ (3)

R: horizontal range of the ball = 20.0 m

You solve the previous equation for vo, the initial horizontal velocity:

$v_o=R\sqrt{\frac{g}{2h}}=(20.0m)\sqrt{\frac{9.8m/s^2}{2(1.60m)}}\\\\v_o=35\frac{m}{s}$

The horizontal component of the velocity is constant in the complete trajectory, hence, you have that

vx = vo = 35 m/s

Finally, you replace the values of vx and vy in the equation (1):

$v=\sqrt{(35m/s)^2+(31.36m/s)^2}=46.99\frac{m}{s}$

The velocity of the ball just before it touches the ground is 46.99 m/s

5 0

3 years ago

What is its maximum altitude above the ground? The answer is the maximum height above the ground

Kisachek [45]

Answer:

Maximum altitude above the ground = 1,540,224 m = 1540.2 km

Explanation:

Using the equations of motion

u = initial velocity of the projectile = 5.5 km/s = 5500 m/s

v = final velocity of the projectile at maximum height reached = 0 m/s

g = acceleration due to gravity = (GM/R²) (from the gravitational law)

g = (6.674 × 10⁻¹¹ × 5.97 × 10²⁴)/(6370000²)

g = -9.82 m/s² (minus because of the direction in which it is directed)

y = vertical distance covered by the projectile = ?

v² = u² + 2gy

0² = 5500² + 2(-9.82)(y)

19.64y = 5500²

y = 1,540,224 m = 1540.2 km

Hope this Helps!!!

3 0

3 years ago

13. You push with 56 N on a 15-kg box, and there is a 23-N force of friction. How fast will the box accelerate?

Flura [38]

Answer:

The acceleration is 2.2 m/s^2

Explanation:

In the attached image, we can see the free body diagram. And using the second law of Newton it will be possible to find the acceleration of the box.

4 0

3 years ago

Three forces act on an object. Two of the forces are at an angle of 100◦to each other and have magnitude 25N and 12N. The third

seraphim [82]

Answer:

F₄ = 29.819 N

Explanation:

Given

F₁ = (- 25*Cos 50° i + 25*Sin 50° j + 0 k) N

F₂ = (12*Cos 50° i + 12*Sin 50° j + 0 k) N

F₃ = (0 i + 0 j + 4 k) N

Then we have

F₁ + F₂ + F₃ + F₄ = 0

⇒ F₄ = - (F₁ + F₂ + F₃)

⇒ F₄ = - ((- 25*Cos 50° i + 25*Sin 50° j) N + (12*Cos 50° i + 12*Sin 50° j) N + (4 k) N) = (13*Cos 50° i - 37*Sin 50° j - 4 k) N

The magnitude of the force will be

F₄ = √((13*Cos 50°)² + (- 37*Sin 50°)² + (- 4)²) N = 29.819 N

6 0

3 years ago

ASAP

scoundrel [369]

Answer:

Explanation:

Hooke's law! F(spring)=-kx

There's no tricky square law here. The spring constant doesn't change, only x (distance stretched) changes. Therefore, if distance is halved, Force will be halved.

5 0

2 years ago