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AVprozaik [17]
3 years ago
6

A block of mass 0.260 kg is placed on top of a light, vertical spring of force constant 5 200 N/m and pushed downward so that th

e spring is compressed by 0.090 m. After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise
Physics
1 answer:
abruzzese [7]3 years ago
4 0

After being released, the restoring force exerted by the spring performs

1/2 (5200 N/m) (0.090 m)² = 12.06 J

of work on the block. At the same time, the block's weight performs

- (0.260 kg) <em>g</em> (0.090 m) ≈ -0.229 J

of work. Then the total work done on the block is about

<em>W</em> ≈ 11.83 J

The block accelerates to a speed <em>v</em> such that, by the work-energy theorem,

<em>W</em> = ∆<em>K</em>   ==>   11.83 J = 1/2 (0.260 kg) <em>v</em> ²   ==>   <em>v</em> ≈ 9.54 m/s

Past the equilibrium point, the spring no longer exerts a force on the block, and the only force acting on it is due to its weight, hence it has a downward acceleration of magnitude <em>g</em>. At its highest point, the block has zero velocity, so that

0² - <em>v</em> ² = -2<em>gy</em>

where <em>y</em> is the maximum height. Solving for <em>y</em> gives

<em>y</em> = <em>v</em> ²/(2<em>g</em>) ≈ 4.64 m

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A 2.4-kg ball falling vertically hits the floor with a speed of 2.5 m/s and rebounds with a speed of 1.5 m/s. What is the magnit
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9.6 Ns

Explanation:

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I = m(v-u).................. Equation 1

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3 years ago
A 115 g hockey puck sent sliding over ice is stopped in 15.1 m by the frictional force on it from the ice.
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Answer:

(a) Ff = 0.128 N

(b μk = 0.1135

Explanation:

kinematic analysis

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d= v₀t+ (1/2)*a*t² Formula (2)

Where:  

d:displacement in meters (m)  

t : time in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s

Calculation of the acceleration of the  hockey puck

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vf=v₀+a*t      v₀=5.8 m/s ,  vf=0

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15.1= 5.8*t-2.9*t

15.1= 2.9*t

t = 15.1 / 2.9

t= 5.2 s

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a= -5.8/5.2

a= -1.115 m/s²

(a) Calculation of the  frictional force (Ff)

We apply Newton's second law

∑F = m*a    Formula (3)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Look at the free body diagram of the  hockey puck in the attached graphic

∑Fx = m*a     m= 115g * 10⁻³ Kg/g = 0.115g    ,  a= -1.12 m/s²

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(b) Calculation of the coefficient of friction (μk)

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g: acceleration due to gravity =9.8 m/s²

∑Fy = 0

N-W=0

N = W

N =  1.127 N

μk = Ff/N

μk = 0.128/1.127

μk = 0.1135

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A)<span>
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D) dE/dr = 0 = 12kQ(1/R³ - r/R^4) means that 
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<span>E) E = 12kQ(R/R³ - R²/(2R^4)) = 12kQ / 2R² = 6kQ / R² </span></span>

4 0
3 years ago
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