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OleMash [197]
2 years ago
9

you push with an 18-N horizontal force on a 5-kg box of coffee resting on a on a horizontal surface. the force of friction on th

e box is 8-N. what is its acceleration and final velocity and its final position if pushed at this rate for 10 seconds
Physics
1 answer:
KatRina [158]2 years ago
3 0

The acceleration is found to be 2 m/s², final velocity after 10 seconds is 20 m/s and the final position after 10 seconds is 100 m away from the starting point.

Answer:

Explanation:

As per Newton's second law of motion, acceleration of any object is directly proportional to the net external unbalanced force acting on that object and inversely proportional to the mass of the object.

Since there are two forces acting on the box in opposite direction, the net force will be the difference of horizontal and frictional force acting on the object.

Net force = Horizontal force - Frictional force = 18 N - 8 N = 10 N.

Now, from second law of motion, Acceleration = \frac{Net force}{Mass}

So, acceleration = 10 N /5 kg = 2 m/s².

Since, acceleration exerted by the box is found to be 2 m/s², we can determine the final velocity of the object after 10 seconds using the first equation of motion.

v = u + at, Here v is the final velocity and u is the initial velocity which is zero for the present case. Other parameters like a is found to be 2 m/s² and time is 10 seconds.

So Final velocity v = 0+(2×10)=20 m/s.

And the final position can be determined using the second equation of motion.

s = ut+1/2at²

Final position = (0×10)+(0.5×2×10×10)= 100 m.

So the final position is 100 m.

Thus, the acceleration is found to be 2 m/s², final velocity after 10 seconds is 20 m/s and the final position after 10 seconds is 100 m away from the starting point.

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Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2} (1)

x=V_{ox}t (2)

V_{f}=V_{oy}-gt (3)

Where:

y=0 m is the bomb's final jeight

y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m is the bomb'e initial height

V_{oy}=0 m/s is the bomb's initial vertical velocity, since the airplane was moving horizontally

t is the time

g=9.8 m/s^{2} is the acceleration due gravity

x is the bomb's range

V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s is the bomb's initial horizontal velocity

V_{f} is the bomb's fina velocity

Knowing this, let's begin with the answers:

<h3>b) Time</h3>

With the conditions given above, equation (1) is now written as:

y_{o}=\frac{1}{2}gt^{2} (4)

Isolating t:

t=\sqrt{\frac{2 y_{o}}{g}} (5)

t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}} (6)

t=17.49 s (7)

<h3>a) Final velocity</h3>

Since V_{oy}=0 m/s, equation (3) is written as:

V_{f}=-gt (8)

V_{f}=-(97.22)(17.49 s) (9)

V_{f}=-171.402 m/s (10) The negative sign ony indicates the direction is downwards

<h3>c) Range</h3>

Substituting (7) in (2):

x=(97.22 m/s)(17.49 s) (11)

x=1700.99 m (12)

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My response to question (a) and (b) is that all of the element of the rope need to  aid or support the weight of the rope and as such, the tension will tend to increase along with height.

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