Answer:
a)[tex]W=0.8J[/tex]
b) [tex]d_t=0.20m[/tex]
c) [tex]\triangle K.E=0.267J[/tex]
Explanation:
From the question we are told that:
Mass of cart 1 [tex]M_1=1.0kg[/tex]
Mass of cart 1 [tex]M_2=0.05kg[/tex]
Force on cart 1 [tex]F_1=4.0N[/tex]
Push Distance of cart 1 [tex]d_1=0.20m[/tex]
Slide Distance of cart 1 [tex]d_1'=0.35m[/tex]
a)
Generally the equation for work-done is mathematically given by
[tex]W=f*d\\W=4*0.20\\W=0.8J \\[/tex]
b)
The systems center of mass moved a net totally of (while being pushed)
Mass 1 =0.20m
Mass 2=0
Therefore
[tex]d_t=d_1+d_2[/tex]
[tex]d_t=0.20+0[/tex]
[tex]d_t=0.20m[/tex]
c)
Since work-done is equal to K.E energy of cart 1
Therefore
[tex]W=1/2mv^2[/tex]
[tex]V_1=\sqrt{\frac{W}{1/2m}}[/tex]
[tex]V_1=\sqrt{\frac{0.8}{1/2(1)}}[/tex]
[tex]V_1=1.264[/tex]
Therefore Kinetic energy before collision is
[tex]K.E_1=1/2mv^2[/tex]
[tex]K.E_1=1/2*1*1.264^2[/tex]
[tex]K.E_1=0.768[/tex]
Generally from the equation for conservation of momentum the Velocity of cart 2 is mathematically given by
[tex]v_2=\frac{m_1V_1}{m_1+m_2}[/tex]
[tex]v_2=\frac{1*1.264}{1+0.5}[/tex]
[tex]V_2=0.842m/s[/tex]
Therefore the final K.E is mathematically given by
[tex]K.E_2=(1/2)(m_1+m_2)V_2^2[/tex]
[tex]K.E_2=1/2*(1.5)(0.842)^2[/tex]
[tex]K.E_2=0.531J[/tex]
Generally the Change in K.E is mathematically given by
[tex]\triangle K.E=K.E_1-K.E_2[/tex]
[tex]\triangle K.E=0.798-0.531[/tex]
[tex]\triangle K.E=0.267J[/tex]
Therefore the will force change the kinetic energy of the system's center of mass by
[tex]\triangle K.E=0.267J[/tex]
(a) The work done by you when you push the cart at a constant force is 0.8 J.
(b) The distance moved by the center mass of the two cart system is 0.23 m.
(c) The change in kinetic energy of the system center of mass is 0.271 J.
The work done by you when you push the cart at a constant force is calculated as follows;
W = Fd
W = 4 x 0.2
W = 0.8 J
Xcm = (m₁x₁ + m₂x₂)/(m₁ + m₂)
Xcm = (0.5(0) + 1(0.35)) / (1 + 0.5)
Xcm = (0.35)/(1.5)
Xcm = 0.23 m
F = ma
a = F/m
a = (4)/1 = 4 m/s²
v² = u² + 2as
v² = 0 + 2(4)(0.2)
v² = 1.6
v = √1.6
v = 1.265 m/s
m₁u₁ + m₂u₂ = v(m₁ + m₂)
1(1.265) + 0 = v(1 + 0.5)
1.265 = 1.5v
v = 0.84 m/s
K.E(initial) = ¹/₂m₁u₁² + ¹/₂m₂u₂²
K.E(initial) = ¹/₂(1)(1.265)² + ¹/₂(0.5)(0) = 0.8 J
K.E(final) = ¹/₂(m₁ + m₂)v²
K.E(final) = ¹/₂(1 + 0.5)(0.84)² = 0.529 J
Δ K.E = 0.8 J - 0.529 J = 0.271 J
Learn more about change in kinetic energy here: https://brainly.com/question/1932411
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