All categories

2 years ago

Please need help with this

Physics

1 answer:

NeTakaya2 years ago

7 0

Note that:

The gravitational potential energy = $mgh$

where m: is the mass, g: the acceleration due to the gravity and h is the height from the earth surface

Then, we can increase the gravitational potential energy by increasing the mass or the height from the earth surface

In our question, we can increase the gravitational potential energy by

A) Strap a boulder to the car so that it wights more.

You might be interested in

A 2.7-kg block is released from rest and allowed to slide down a frictionless surface and into a spring. The far end of the spri

exis [7]

a) The speed of the block at a height of 0.25 m is 2.38 m/s

b) The compression of the spring is 0.25 m

c) The final height of the block is 0.54 m

Explanation:

We can solve the problem by using the law of conservation of energy. In fact, the total mechanical energy (sum of kinetic+gravitational potential energy) must be conserved in absence of friction. So we can write:

$U_i +K_i = U_f + K_f$

where

$U_i$ is the initial potential energy, at the top

$K_i$ is the initial kinetic energy, at the top

$U_f$ is the final potential energy, at halfway

$K_f$ is the final kinetic energy, at halfway

The equation can be rewritten as

$mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2$

where:

m = 2.7 kg is the mass of the block

$g=9.8 m/s^2$ is the acceleration of gravity

$h_i = 0.54$ is the initial height

u = 0 is the initial speed

$h_f = 0.25 m$ is the final height of the block

v is the final speed when the block is at a height of 0.25 m

Solving for v,

$v=\sqrt{u^2+2g(h_i-h_f)}=\sqrt{0+2(9.8)(0.54-0.25)}=2.38 m/s$

The total mechanical energy of the block can be calculated from the initial conditions, and it is

$E=K_i + U_i = 0 + mgh_i = (2.7)(9.8)(0.54)=14.3 J$

At the bottom of the ramp, the gravitational potential energy has become zero (because the final heigth is zero), and all the energy has been converted into kinetic energy. However, then the block compresses the spring, and the maximum compression of the spring occurs when the block stops: at that moment, all the energy of the block has been converted into elastic potential energy of the spring. So we can write

$E=E_e = \frac{1}{2}kx^2$

where

k = 453 N/m is the spring constant

x is the compression of the spring

And solving for x, we find

$x=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(14.3)}{453}}=0.25 m$

If there is no friction acting on the block, we can apply again the law of conservation of energy. This time, the initial energy is the elastic potential energy stored in the spring:

$E=E_e = 14.3 J$

while the final energy is the energy at the point of maximum height, where all the energy has been converted into gravitational potetial energy:

$E=U_f = mg h_f$

where $h_f$ is the maximum height reached. Solving for this quantity, we find

$h_f = \frac{E}{mg}=\frac{14.3}{(2.7)(9.8)}=0.54 m$

which is the initial height: this is correct, because the total mechanical energy is conserved, so the block must return to its initial position.

Learn more about kinetic and potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

brainly.com/question/6536722

#LearnwithBrainly

5 0

3 years ago

What is static and sliding friction

vodka [1.7K]

Static friction is the friction that exists between a stationary object and the surface on which it's resting.
frictional force occurs when you try to push an object alongside a surface.

5 0

2 years ago

g A 2.3 kg block is attached to the spring, and it is released from rest 0.7 m from the spring's equilibrium position. Neglectin

DedPeter [7]

Answer:

3.71 m/s

Explanation:

From the law of conservation of linear momentum, since we are neglecting minor energy losses due to friction then we can express it as $mgh=0.5mv^{2}$ since all the potential energy is transformed to kinetic energy

Making v the subject of the formula then $v=\sqrt {2gh}$ and here m is the mass of the block, g is acceleration due to gravity, h is the height. Substituting 0.7 m for h and 9.81 for g then we obtain that $v=\sqrt {2\times 9.81\times 0.7}=3.705941176 m/s\approx 3.71 m/s$

6 0

3 years ago

What is a single cell organism able to do?

Sunny_sXe [5.5K]

Survived on its own.

7 0

3 years ago

Read 2 more answers

9. Captain America is chasing Red Skull. He plans to throw his shield to knock down Red Skull but needs to know how fast Red Sku

Sedaia [141]

Red Skull's relative velocity to Captain America, towards the left front of the

truck is approximately 33.23 m/s in a direction from the North of

approximately 9.18°.

Reasons:

Assumptions;

Taking the north direction as positive.

The activity takes place on the trucks.

The trucks are moving towards each other.

Solution:

Vector form of net speed of Red Skull, is given as follows;

v₁ = -( $\frac{\sqrt{2} }{2}$ × 3.5)·i + ( $\frac{\sqrt{2} }{2}$ × 3.5 + 12.5)·j

Vector form of the net speed of Captain America is given as follows;

v₂ = ( $\frac{\sqrt{2} }{2}$ × 4.0)·i - ( $\frac{\sqrt{2} }{2}$ × 4.0 + 15)·j

Relative velocity, v₁₂ = v₁ - v₂

∴ v₁₂ = (-( $\frac{\sqrt{2} }{2}$ × 3.5) - ( $\frac{\sqrt{2} }{2}$ × 4.0))·i + (( $\frac{\sqrt{2} }{2}$ × 3.5 + 12.5) + ( $\frac{\sqrt{2} }{2}$ × 4.0 + 15))·j

v₁₂ = $-\frac{ 15 \cdot \sqrt{2} }{4}$ ·i + $\frac{ 110 + 15 \cdot \sqrt{2} }{4}$ ·j

Red Skull's velocity relative to Captain America, v₁₂ = $-\frac{ 15 \cdot \sqrt{2} }{4}$ ·i + $\frac{ 110 + 15 \cdot \sqrt{2} }{4}$ ·j

v₁₂ ≈ -5.3·i + 32.8·j

Therefore;

Red Skull appears to be moving West at 5.3 m/s and North at 32.8 m/s

The direction is $arctan \left(\frac{32.8}{-5.3} \right) \approx -80.2^{\circ}$

Therefore;

Red Skull appear to be moving at 90° - 80.2° ≈ 9.18° towards the left front end of the truck moving North

The magnitude of the velocity, |v₁₂|, is given as follows;

$|v_{12}| = \sqrt{\left(-\frac{ 15 \cdot \sqrt{2} }{4}\right)^2 + \left(\frac{ 110 + 15 \cdot \sqrt{2} }{4}\right)^2} = \dfrac{ 5 \cdot \sqrt{130+33 \cdot\sqrt{2} } }{2} \approx 33.23$ ·

The magnitude of Red Skull's velocity relative to Captain America is,

therefore;

|v₁₂| ≈ 33.23 m/s

Learn more here:

brainly.com/question/24430414

6 0

2 years ago