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ch4aika [34]
3 years ago
13

A web page designer creates an animation in which a dot on a computer screen has a position of r⃗ =[ 4.50 cm +( 2.90 cm/s2 )t2]i

^+( 4.70 cm/s )tj^ . Part A Find the average velocity of the dot between t=0 and t=2.0s. Give your answer as a pair of components separated by a comma. For example, if you think the x component is 3 and the y component is 4, then you should enter 3,4.
Physics
1 answer:
VMariaS [17]3 years ago
5 0

Answer:

V = (5.8cm/s)i, (4.7cm/s)j

Explanation:

Given :

r⃗ =[ 4.50 cm +( 2.90 cm/s2 )t2]i^+( 4.70 cm/s )tj^

To obtain the average velocity (V)

V = (r2 - r1) / (t2 - t1)

To obtain r1 and r2, substitute t1 = 0 and t2 = 2 respectively in the equation above

r1 = [ 4.50 cm +( 2.90 cm/s2 ) 0]i^+( 4.70 cm/s )0 j

r1 = 4.50 cm + 0 + 0 = (4.50cm)i + 0j

r2 = [ 4.50 cm +( 2.90 cm/s2 )2²]i^+( 4.70 cm/s )2 j

r2 = 4.50cm + (2.90 × 4)i + (4.70 × 2)j

r2 = (16.1cm)i + (9.4cm)j

V = [(16.1 - 4.50)i - (9.4 - 0)j] / 2 - 0

V = 11.6i / 2 ; 9.4j / 2

V = (5.8cm/s)i, (4.7cm/s)j

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A 50.0-g object connected to a spring with a force constant of 35.0 n/m oscillates with an amplitude of 4.00 cm on a frictionles
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A) The total energy of the system is sum of kinetic energy and elastic potential energy:
E=K+U= \frac{1}{2}mv^2 +  \frac{1}{2}kx^2
where
m is the mass
v is the speed
k is the spring constant
x is the elongation/compression of the spring

The total energy is conserved, so we can calculate its value at any point of the motion. If we take the point of maximum displacement:
x=A=4.00 cm = 0.04 m
then the velocity of the system is zero, so the total energy is just potential energy, and it is equal to
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b) When the position of the object is 
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the potential energy of the system is
U= \frac{1}{2}kx^2 =  \frac{1}{2}(35.0 N/m)(0.01 m)^2 = 1.75 \cdot 10^{-3} J
and so the kinetic energy is
K=E-U=0.028 J - 1.75 \cdot 10^{-3}J =0.026 J
since the mass is m=50.0 g=0.05 kg, and the kinetic energy is given by
K= \frac{1}{2}mv^2
we can re-arrange the formula to find the speed of the object:
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c) The potential energy when the object is at 
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U= \frac{1}{2}kx^2 =  \frac{1}{2}(35.0 N/m)(0.03 m)^2 =0.016 J
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K=E-U=0.028 J-0.016 J = 0.012 J

d) We already found the potential energy at point c, and it is given by
U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.03 m)^2 =0.016 J
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