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Vsevolod [243]
2 years ago
8

You attach a 1.10 kg block to a horizontal spring that is fixed at one end. You pull the block until the spring is stretched by

0.200 m and release it from rest. Assume the block slides on a horizontal surface with negligible friction. The block reaches a speed of zero again 0.100 s after release (for the first time after release). What is the maximum speed of the block (in m/s)?
Physics
1 answer:
dangina [55]2 years ago
6 0

Answer:

6.3 m/s

Explanation:

m = mass of the block = 1.10 kg

k = spring constant of the spring

x = stretch in the spring = 0.2 m

t = time taken by block to come to zero speed first time = 0.100 s

T = Time period of oscillation

Time period of oscillation is given as

T = 2t

T = 2 (0.1)

T = 0.2 s

Time period is also given as

T = 2\pi \sqrt{\frac{m}{k}}

0.2 = 2(3.14) \sqrt{\frac{1.10}{k}}

k = 1084.6 N/m

v = maximum speed of the block

using conservation of energy

Maximum kinetic energy = Maximum spring potential energy

(0.5) m v² = (0.5) k x²

m v² = k x²

(1.10) v² = (1084.6) (0.2)²

v = 6.3 m/s

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Six dogs pull a two-person sled with a total mass of 220 kg. The coefficient of kinetic friction between the sled and the snow i
irina [24]

Answer:

a) 2457J

b) 558W

c) 337N

Explanation:

Assuming dogs started from rest.

v=a*t\\t=\frac{a}{v}\\v=12\frac{km}{h}\frac{1000m}{km}*\frac{1h}{3600s}\\\\v=3.3m/s\\\\t=\frac{3.3m/s}{0.75m/s^2}\\t=4.4s

and the displacement is given by:

d=\frac{1}{2}*a*t^2\\d=7.3m

Using the energy conservation formula:

K_i+U_i+W_d+W_f=K_f+U_f

Because the motion started from rest the initial kinetic energy is zero, the motion occurred in-ground level so the gravitational energy is zero too.

the work done by the friction force is given by:

W_f=F_f*d*cos(\theta)\\W_f=\µ*m*g*d*cos(180)\\W_f=0.08*220kg*9.8m/s^2*7.3m*(-1)\\W_f=-1259J

so:

W_d=\frac{1}{2}*220kg*(3.3m/s)^2+1259J\\W_d=2457J

The power is given by:

P=\frac{W}{t}\\\\P=\frac{2457J}{4.4s}\\\\P=558W

and the force exerted by the dogs:

W_d=F_d*d*cos(\theta)\\F_d=\frac{W_d}{d*cos(0)}\\\\F=\frac{2457J}{7.3m*(1)}\\\\F=337N

4 0
2 years ago
If the loop the car is currently on has a radius of 26.0 m , find the minimum height h so that the car will not fall off the tra
ki77a [65]

The minimum height h is 65m so that the car will not fall off the track at the top of the circular part of the loop.

<h3>What is mechanical energy?</h3>

Potential energy plus kinetic energy are combined to form mechanical energy. According to the principle of mechanical energy conservation, mechanical energy is constant in an isolated system when only conservative forces are acting on it. Potential energy increases when an object moves in the opposite direction of a conservative net force. Kinetic energy also changes as an object's speed, not velocity, changes. However, nonconservative forces, such as frictional forces, will always be present in real systems; however, if these forces are of minimal magnitude, mechanical energy changes little, making the idea of its conservation a reasonable approximation.

For completing the vertical circle the minimum speed at the bottom must be \sqrt{5gR}

so conserving mechanical energy

mgh=\frac{1}{2} m (v_{bottom})^{2}

gh=\frac{1}{2} 5gR

⇒ h= \frac{5}{2} \times 26

h = 65m

To learn more about mechanical energy, visit:

brainly.com/question/24443465

#SPJ4

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1 year ago
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