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Vsevolod [243]
3 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]3 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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USPshnik [31]

Answer:

ΔK = -6 10⁴ J

Explanation:

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initial instant. Before the crash

        p₀ = m v₁ + M 0

final instant. Right after the crash

        p_f = (m + M) v

        p₀ = p_f

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        v = \frac{m}{m+M} \  v_1

     

we substitute

        v = \frac{20}{20+40}   3

        v = 1.0 m / s

having the initial and final velocities, let's find the kinetic energy

        K₀ = ½ m v₁² + 0

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        K_f = ½ (m + M) v²

        K_f = ½ (20 +40) 10³  1²

        K_f = 3 10⁴ J

the change in energy is

       ΔK = K_f - K₀

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The negative sign indicates that the energy is ranked in another type of energy

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The stair stepper is a novel exercise machine that attempts to reproduce the work done against gravity by walking up stairs. Wit
nika2105 [10]

Answer:

The total work done by Brad each day is 176400 J

Explanation:

Hi there!  The work done by a force (F) pointed in the same direction as the displacement (d) is calculated as follows:

W = F · d

The force applied is equal to the weight of Brad, that is calculated as follows:

Weight = m · g

Where:

m =  mass of Brad

g = acceleration due to gravity (9.8 m/s²)

Then:

Weight = 60 kg · 9.8 m/s² = 588 N

 

Let´s find the vertical distance traveled by Brad each day:

He exercises 20 min per day. Each minute Brad does 60 steps. In total, Brad steps up (20 min · 60 steps/min) 1200 steps. If each step has a height of 0.25 m, the total distance traveled by Brad will be

(1200 steps · 0.25 m/step) 300 m.

Then, the total work done by Brad is

W = F · d

W =  588 N · 300 m

W = 176400 J

The total work done by Brad each day is 176400 J

6 0
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