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3 years ago

• How much work isrequired to lift a 2kgobject 2m high?

1 answer:

7 0

Answer You need to consider that the gravity on earth is 9.8 m/s/s. This means any object you let go on the earths surface will gain 9.8 m/s of speed every second. You need to apply a force on the object in the opposite direction to avoid this acceleration. If you are pushing something up at a constant speed, you are just resisting earths acceleration. The more massive and object is, the greater force is needed to accelerate it. The equation is Force = mass*acceleration. So for a 2kg object in a 9.8 m/s/s gravity you need 2kg*9.8m/s/s = 19.6 Newtons to counteract gravity. Work or energy = force * distance. So to push with 19.6 N over a distance of 2 meters = 19.6 N*2 m = 39.2 Joules of energy. There is an equation that puts together those two equations I just used and it is E = mgh

The amount of Energy to lift an object is (mass) * (acceleration due to gravity) * (height)

:Hence, the Work done to life the mass of 2 kg to a height of 10 m is 196 J. Hope it helps❤️❤️❤️

Explanation:

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2 years ago

The box is pushed to the right with a force of 40 Newtons and it just begins to move what is the maximum static frictional force

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Answer:

The maximum static frictional force is 40N.

Explanation:

When an object of mass M is on a surface with a coefficient of static friction μ, there is a minimum force that you need to apply to the object in order to "break" the coefficient of static friction and be able to move the object (Called the threshold of motion, once the object is moving we have a coefficient of kinetic friction, which is smaller than the one for static friction).

This coefficient defines the maximum static friction force that we can have.

So if we apply a small force and we start to increase it, the static frictional force will be equal to our force until it reaches its maximum, and then we can move the object and now we will have frictional force.

In this case, we know that we apply a force of 40N and the object just starts to move.

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3 years ago

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. choose the origin to be at the location where the bullet begin

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Part a) The work done by the gas on the bullet is the integral of the force in dx, where x is the distance covered by the bullet inside the barrel with respect to the origin:
$W= \int\limits^{0.540m}_{0} {F} \, dx = \int\limits^{0.540m}_{0} {(16000+10000x-26000x^2)} \, dx =$
$=16000x+10000 \frac{x^2}{2} - 26000 \frac{x^3}{3}$
By substituting the length of the barrel, L=0.540 m, we find the total work done by the gas on the bullet:
$W=16000(0.540m)+10000 \frac{(0.540m)^2}{2} - 26000 \frac{(0.540m)^3}{3} =$
$=8733 J=8.73 kJ$

part b) The resolution of the problem is the same, we just have to use the new length of the barrel (L=0.95 m) inside the final formula, and we find the new value of the work:
$W=16000(0.95m)+10000 \frac{(0.95m)^2}{2} - 26000 \frac{(0.95m)^3}{3} =$
$=12280 J=12.28 kJ$

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3 years ago

The regions of the Moon that are of higher elevation are known as the __________. These regions have a high albedo and display a

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Answer:

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The lunar highlands are seen from the earth a <u>whiter color than the rest of the moon</u> due to its altitude.

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4 years ago

• How much work isrequired to lift a 2kgobject 2m high?​

• How much work isrequired to lift a 2kgobject 2m high?