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OLga [1]
3 years ago
10

Two blocks A and B with mA = 2.6 kg and mB = 0.81 kg are connected by a string of negligible mass. They rest on a frictionless h

orizontal surface. You pull on block A with a horizontal force of 6.5 N. Determine the tension in the string connecting the two blocks.
Physics
1 answer:
vladimir1956 [14]3 years ago
5 0

Answer:

Explanation:

Given

Mass of block A (m_a)=2.6 kg

Mass of block B (m_b)=0.81 kg

Force of 6.5 N is applied on Block A

Force on Block A is F, Tension

thus

F-T=m_a\times a -----1

where a=acceleration of the system

For Block B

T=m_b\times a -----2

F=m_b\times a+m_a\times a

F=\left ( m_a+m_b\right )a

F=\left ( 2.6+0.81\right )a

a=\frac{6.5}{3.41}

a=1.96 m/s^2

T=0.81\times 1.96=1.54 N

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A small meteorite with mass of 1 g strikes the outer wall of a communication satellite with a speed of 2Okm/s (relative to the s
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The energy coverted to heat is 200 kilojoules.

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GIven the absence of external forces exerted both on the small meteorite and on the communication satellite, the Principle of Linear Momentum is considered and let suppose that collision is completely inelastic and that satellite is initially at rest. Hence, the expression for the satellite-meteorite system:

m_{M}\cdot v_{M} + m_{S}\cdot v_{S} = (m_{M}+m_{S})\cdot v

Where:

m_{M}, m_{S} - Masses of the small meteorite and the communication satellite, measured in kilograms.

v_{M}, v_{S} - Speeds of the small meteorite and the communication satellite, measured in meters per second.

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The final speed of the satellite-meteorite system is cleared:

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v = \frac{(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)}{1\times 10^{-3}\,kg+200\,kg}

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Which means that the new system remains stationary and all mechanical energy from meteorite is dissipated in the form of heat. According to the Principle of Energy Conservation and the Work-Energy Theorem, the change in the kinetic energy is equal to the dissipated energy in the form of heat:

K_{S} + K_{M} - K - Q_{disp} = 0

Q_{disp} = K_{S}+K_{M}-K

Where:

K_{S}, K_{M} - Initial translational kinetic energies of the communication satellite and small meteorite, measured in joules.

K - Kinetic energy of the satellite-meteorite system, measured in joules.

Q_{disp} - Dissipated heat, measured in joules.

The previous expression is expanded by using the definition for the translational kinetic energy:

Q_{disp} = \frac{1}{2}\cdot [m_{M}\cdot v_{M}^{2}+m_{S}\cdot v_{S}^{2}-(m_{M}+m_{S})\cdot v^{2}]

Given that m_{M} = 1\times 10^{-3}\,kg, m_{S} = 200\,kg, v_{M} = 20000\,\frac{m}{s}, v_{S} = 0\,\frac{m}{s} and v = 0.1\,\frac{m}{s}, the dissipated heat is:

Q_{disp} = \frac{1}{2}\cdot \left[(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)^{2}+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)^{2}-(200.001\,kg)\cdot \left(0.001\,\frac{m}{s} \right)^{2}\right]Q_{disp} = 200000\,J

Q_{disp} = 200\,kJ

The energy coverted to heat is 200 kilojoules.

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