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3 years ago

Kinetic energy help! (#4)

Physics

1 answer:

zloy xaker [14]3 years ago

5 0

For problem number 4:

It says it is moving 10% of the speed of light. The speed of light is 3*10^8 m/s.
So:

KE = 0.5mv^2 = 0.5*(1.67*10^-27kg)(3*10^8*0.1)^2
KE = 7.515*10^-13 J

Problem 5:
W = ΔKE
Since it is initially at rest, we can simplify W = KE. KE = 0.5mv^2

3.4*10^-16J = 0.5(9.11*10^-31)v^2
v = 27320916.74 m/s

Problem 6:
mgh = 0.5mv^2
(0.1kg)*(10m/s^2)*(16m) = 0.5(0.1kg)v^2
v = 17.8885m/2

Did this really quickly, so might not be accurate!

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The value is $A = 0.014 \ m$

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From the question we are told that

The mass of the object is $m = 2.0 \ kg$

The unstressed length of the string is $l = 0.08 \ m$

The length of the spring when it is at equilibrium is $l_e = 5.9 \ cm = 0.059 \ m$

The initial speed (maximum speed)of the spring when given a downward blow $v = 0.30 \ m/s$

Generally the maximum speed of the spring is mathematically represented as

$u = A * w$

Here A is maximum height above the floor (i.e the maximum amplitude)

and $w$ is the angular frequency which is mathematically represented as

$w = \sqrt{\frac{k}{m} }$

$u = A * \sqrt{\frac{k}{m} }$

=> $A = u * \sqrt{\frac{m}{k} }$

Gnerally the length of the compression(Here an assumption that the spring was compressed to the ground by the hammer is made) by the hammer is mathematically represented as

$b = l -l_e$