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Reil [10]
3 years ago
15

Two forces act in opposite directions on a wood block. What will happen if the forces are unbalanced?

Physics
1 answer:
mihalych1998 [28]3 years ago
8 0

Answer: Its the last one: The position of the block will change

Explanation: Since the forces are unbalanced, one side of the block will be stronger, therefore making the block move.

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Use the values from PRACTICE IT to help you work this exercise. Suppose the same two vehicles are both traveling eastward, the c
Mariulka [41]

Answer:

A. v_{3}=12.17m/s

B. v_{car}=6.3m/s\\v_{truck}=-6.3m/s

C. ΔK=-4.13x10^3J

Explanation:

From the exercise we know that the car and the truck are traveling eastward. I'm going to name the car 1 and the truck 2

v_{1}=5.79m/s\\m_{1}=102kg\\v_{2}=18.5m/s\\m_{2}=103kg

A. Since the two vehicles become entangled the final mass is:

m_{3}=102kg+103kg=205kg

From linear momentum we got that:

p_{1}=p_{2}

m_{1}v_{1}+m_{2}v_{2}=m_{3}v_{3}

v_{3}=\frac{m_{1}v_{1}+m_{2}v_{2}}{m_{3} }=\frac{(102kg)(5.79m/s)+(103kg)(18.5m/s)}{(205kg)}

v_{3}=12.17m/s

B. The change in velocity of both vehicles are:

For the car

v_{car}=v_{f}-v_{o}=12.17m/s-5.79m/s=6.38m/s

For the truck

v_{truck}=12.17m/s-18.5m/s=-6.3m/s

C. The change in kinetic energy is:

ΔK=K_{2}-K_{1} =\frac{1}{2}m_{3}v_{3}^{2}-(\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2})

ΔK=\frac{1}{2}(205)(12.17)^{2}-(\frac{1}{2}(102)(5.79)^{2}+\frac{1}{2}(103)(18.5)^{2})=-4.13x10^{3}J

ΔK=-4.13x10^{3}J

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