It takes 100 n to move a box 10 meters in 5 seconds. how much power is required?

Physics

2 answers:

mylen [45]3 years ago

8 0

Answer: 200 w in apex

I am Lyosha [343]3 years ago

6 0

<span>Power = Work / Time
Work = Force * Distance

Force is measured in Newtons, so 100 N is the force in this situation.
Distance in this scenario is 10 m. And time is 5 s.

Inserting known values into the formulas:
Work = 100 N * 10 m = 1000 J
(J = Joules, the equivalent of "Newton-meters")

Power = 1000 J / 5 s = 200 W
(W = Watts, the equivalent of 1 Joule / second)

In summary, 200 W of power are required for this situation.</span>

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A projectile is launched at an angle of 29 degrees above the horizontal with an initial velocity of 36.6 at an unknown height.

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A 1.8-kg object is attached to a spring and placed on frictionless, horizontal surface. A force of 40 N stretches a spring 20 cm

Sergio [31]

Answer:

a) k = 200 N/m

b) E = 4 J

c) Δx = 6.3 cm

Explanation:

In order to find force constant of the spring, k, we can use the the Hooke's Law, which reads as follows:

$F = - k * \Delta x (1)$

where F = 40 N and Δx =- 0.2 m (since the force opposes to the displacement from the equilibrium position, we say that it's a restoring force).
Solving for k:

$k =- \frac{F}{\Delta x} =-\frac{40 N}{-0.2m} = 200 N/m (2)$

Assuming no friction present, total mechanical energy mus keep constant.
When the spring is stretched, all the energy is elastic potential, and can be expressed as follows:

$U = \frac{1}{2}* k* (\Delta x)^{2} (3)$

Replacing k and Δx by their values, we get:

$U = \frac{1}{2}* k* (\Delta x)^{2} = \frac{1}{2}* 200 N/m* (0.2m)^{2} = 4 J (4)$

When the object is oscillating, at any time, its energy will be part elastic potential, and part kinetic energy.
We know that due to the conservation of energy, this sum will be equal to the total energy that we found in b).
So, we can write the following expression:

$\frac{1}{2}* k* \Delta x_{1} ^{2} + \frac{1}{2} * m* v^{2} = \frac{1}{2}*k*\Delta x^{2} (5)$