A wooden block with mass 1.05 kg is placed against a compressed spring at the bottom of a slope inclined at an angle of 35.0 degrees (point A). When the spring is released, it projects the block up the incline. At point B, a distance of 4.90m up the incline from A, the block is moving up the incline at a speed of 5.10 m/s and is no longer in contact with the spring. The coefficient of kinetic friction between the block and incline is 0.55. The mass of the spring is negligible. Calculate the amount of potential energy that was initially stored in the spring.

Answers

Answer 1

Answer:

The amount of potential energy that was initially stored in the spring due to the wooden block is 65.3 joules.

What is potential energy?

Potential energy is the energy which body posses because of its position.

The potential energy of a body is given as,

$PE=mgh$

Here, (m) is the mass of the body, (g) is the gravitational force and (h) is the height of the body.

The energy stored in the spring is the sum of all the potential energy, kinetic energy and the energy dissipated due to friction. Therefore, it can be given as,

$E=mgh+(1)/(2)mv^2+\mu mgd\cos\theta$

Here, the mass of the wooden block is 1.05 kg . Angle of inclination is 35.0 degrees (point A). The distance from point B is 4.90m up the incline from A.

The speed of the block is 5.10 m/s and the coefficient of kinetic friction between the block and incline is 0.55. Therefore, put the values in the above formula as,

$E=1.05(9.81)(4.9\sin(35))+(1)/(2)(1.05)(5.1)^2+(0.55)(1.05)(9.8)(4.9)\cos(35)\nE=65.3\rm J$

Hence, the amount of potential energy that was initially stored in the spring due to the wooden block is 65.3 joules.

Learn more about the potential energy here;

brainly.com/question/15896499

Answer 2

Answer:

Answer:

Explanation:

energy stored in spring initially

= kinetic + potential energy of block + energy dissipated by friction

= 1/2 mv² + mgh + μ mgcosθ x d

m is mass , v is velocity at top position , h is vertical height , μ is coefficient of friction ,θ is angle of inclination of plane

= m (1/2 v² + gh + μ gcosθ x d )

= 1.05 ( .5 x 5.1² + 9.8 x 4.9 sin35 + .55 x 9.8 cos35 x 4.9 )

= 1.05 ( 13.005 + 27.543 + 21.635)

= 65.3 J .

Related Questions

Which of the following statements correctly compares the relationship between the earth, its atmosphere and radiation?1. The earth is cooled and its atmosphere is heated by solar radiation. 2. The earth is heated and its atmosphere is cooled by terrestrial radiation. 3. The earth is cooled and its atmosphere is heated by terrestrial radiation. 4. The earth is heated and its atmosphere is cooled by solar radiation. Don't answer unless you know for sure. Thank you so much!
A conveyor belt is used to move sand from one place to another in a factory. The conveyor is tilted at an angle of 18° above the horizontal and the sand is moved without slipping at the rate of 2 m/s. The sand is collected in a big drum 5 m below the end of the conveyor belt. Determine the horizontal distance between the end of the conveyor belt and the middle of the collecting drum.
A typical automobile under hard braking loses speed at a rate of about 7.2 m/s2; the typical reaction time to engage the brakes is 0.55 s. A local school board sets the speed limit in a school zone such that all cars should be able to stop in 3.6 m. What maximum speed does this imply for a car in the school zone?
Which exerts more force, the Earth pulling on the moon or the moon pulling on the Earth? Explain.
Tarik winds a small paper tube uniformly with 163 turns of thin wire to form a solenoid. The tube's diameter is 6.13 mm and its length is 2.49 cm . What is the inductance, in microhenrys, of Tarik's solenoid?

Which volcanoes are formed by pyroclastic deposits? Select all that apply. A. cinder cone B. stratovolcano C. shield volcano

Answers

Answer:

Its a cinder cone cause after it all falls down to make deposits.

cinder cone and stratovolcano

$\huge{\gray{\sf Question:} }$ A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is (a) 5 cm, (b) 3 cm and (c) 0 cm.

Answers

Answer:

$here \: amplitude = 5cm (convert \: to \: si \: unit) = .05m \n t = .2sec \n omega = (2\pi)/(t) \n = (2\pi)/(.2 ) = 10\pi (rad)/(s) \n we \: want \: find \: a \: and \: v \n we \: know \: that \: a = - {omega}^(2) x \n v = omega \sqrt{ {r}^(2) - {x}^(2) } \n(1)x = .05m \n a = - {10\pi}^(2) * .05 = - 5 {\pi}^(2) \frac{m}{ {s}^(2) } \n v = 10\pi \sqrt{ {.05}^(2) - {.05}^(2) } = 0 \n 2)x = 3cm = .03m \n a = {(10\pi)}^(2) * .03 = - 3 {\pi}^(2) \frac{m}{ {s}^(2) } \n v = 10\pi \sqrt{ {.05}^(2) - {.03}^(2) } = 10\pi * .04 = .4\pi (m)/(s) \n 3)x = 0 \n a = - {(10\pi)}^(2) * 0 = 0 \n v = 10\pi * \sqrt{ {.05}^(2) - {0}^(2) } = - 10\pi * .05 = .5\pi (m)/(s) \n thank \: you$

Answer:

ANSWER

A = 5 cm = 0.05 m

T = 0.2 s

ω=2π/T=2π/0.2=10πrad/s

plz give brainlist

hope this helped

A galilean telescope adjusted for a relaxed eye is 36cm long. If the objective lens has a focal length of 40cm, what is the magnification?

Answers

For this problem, we use the mirror equation which is expressed as:

1/di + 1/f = 1/d0

Magnification is expressed as the ratio of di and d0.

Manipulating the equation, we will have:

M = di/f +1
M = 36/40 + 1
M = 1.9

Hope this answers the equation.

A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of the building. Ignore air resistance. (a) How long is the ball in the air?
(b) What must have been the initial horizontal component of the velocity?
(c) What is the vertical component of the velocity just before the ball hits the ground?
(d) What is the velocity (including both the horizontal and vertical components) of the ball just before it hits the ground?

Answers

Answer:

Explanation:

Given

height of building (h)=60 m

Range of ball=100 m

(a)time travel to cover a vertical distance of 60 m

$h=ut+(at^2)/(2)$

$60=0+(9.8* t^2)/(2)$

$t^2=12.24$

t=3.49 s

(b)To cover a range of 100 m

R=ut

$100=v_x* 3.49$

$v_x=28.57 m/s$

(c)vertical component of velocity just before it hits the ground

$v_y=u+at$

$v_y=0+9.81* 3.49=34.202 m/s$

(d) $v_(net)=√(v_y^2+v_x^2)$

$v_(net)=√(1986.02)=44.56 m/s$

Final answer:

The ball is in the air for about 3.5 seconds. The initial horizontal velocity would have been approximately 28.6 m/s. The vertical component of the velocity just before the ball hits the ground is nearly 34.3 m/s. The overall velocity of the ball just prior to impact is roughly 44.6 m/s.

Explanation:

The problem given is about projectile motion which can be approached by splitting the motion into the horizontal and vertical components. We can work out the durations for each.

For the time the ball is in the air, we know that it falls vertically under gravity. Use the equation of motion, h = 0.5gt^2 where h is the height (60.0 m) and g is the acceleration due to gravity (approx 9.81 m/s^2). Solving for t takes approximately 3.5 seconds.
The initial horizontal component of the velocity can be calculated by the distance it traveled horizontally divided by the time it spent in the air. So, 100 m / 3.5 s = 28.6 m/s.
The vertical component of velocity just before the ball hits the ground can be calculated using v = gt where g is the acceleration due to gravity and t is the time. Solving gives around 34.3 m/s.
The overall velocity can then be calculated using the Pythagorean theorem, combining the horizontal and vertical components of the velocity - sqrt((28.6 m/s)^2 + (34.3 m/s)^2). This will approximately equal 44.6 m/s.

Learn more about Projectile Motion here:

brainly.com/question/29545516

#SPJ12

Which formula can be used to calculate the horizontal displacement of a horizontally launched projectile?x = vi(cos )
x = vi(cos )t
x = ayt
x = vxt (RIGHT ANSWER)

Answers

The formula for calculating the horizontal displacement of a horizontally launched projectile is $x=v*t$

A projectile launched horizontally with a velocity v, at a height y ,travels a horizontal distance x, while falling through a distance y. The horizontal velocity of a projectile remains constant throughout its motion, in the absence of air resistance. The vertical component of the velocity is under the action of the gravitational force and hence it increases in magnitude as it falls through the height.

The horizontal displacement of the projectile is a uniform motion and it occurs at a constant speed v.

Thus, the horizontal displacement of the projectile is given by the expression.

$x=v*t$

In a region where there is a uniform electric field that is upward and has magnitude 3.80x104 N/C a small object is projected upward with an initial speed of 2.32 m/s The object travels upward a distance of 5.98 cm in 0 200 s. What is the object's charge-to-mass ratio q/m (magnitude and sign)?Assume g 9.80 m/s and ignore air resistance E3? C/kg q/m