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VARVARA [1.3K]
3 years ago
12

Forces of 11.9 N north, 19.1 N east, and 14.4 N south are simultaneously applied to a 3.77 kg mass as it rests on a frictionless

air table. What is the magnitude of its acceleration?
Physics
1 answer:
oksian1 [2.3K]3 years ago
4 0
Force Vector = <19.1i+11.9j-14.4j>N
Force Vector = <19.1i-2.5j>N
Acceleration Vector =(Force Vector)/mass
Acceleration Vector =<19.1i-2.5j>N/3.77
Acceleration Vector =<5.0663i-0.6631j>m/s^2
|a| = (5.0663^2+(-0.6631)^2)^0.5
|a| = 5.11 (m/s^2)
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The original kinetic energy will be 0 J and the final kinetic energy will be 7500 J and the amount of work utilized will be similar to the final kinetic energy i.e., 7500 J.

<u>Explanation:</u>

As it is known that the kinetic energy is defined as the energy exhibited by the moving objects. So the kinetic energy is equal to the product of mass and square of the velocity attained by the car. Thus,

                  \text {Kinetic energy}=\frac{1}{2} m v^{2}

So the initial kinetic energy will be the energy exerted by the car at the initial state when the initial velocity is zero. Thus the initial kinetic energy will be zero.  

The final kinetic energy is

\text {Kinetic energy}=\frac{1}{2} m v^{2}=\frac{1}{2} \times 600 \times 5 \times 5 = 7500 J

As the work done is the energy required to start the car from zero velocity to 5 m/s velocity.  

                       Work done = Final Kinetic energy - Initial Kinetic energy

Thus the work utilized for moving the car is  

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Thus, the initial kinetic energy of the car is zero, the final kinetic energy is 7500 J and the work utilized by the car is also 7500 J.

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