Someone can lift up to 80kg in a lift that accelerates and descends with an acceleration of 2 5m s2

1) Because A is downward, the person is in a state of weightlessness.

f=m(g-a)

600nf=gm=10m=600n

m=6kg2) Because A is upward at this time, the person is overweight.

f=m'(g+a')=40(10+a')=600na'=5m/s^2

A person can lift up to 80 kg in an elevator that is descending with a uniform acceleration at an accelerated rate, and up to a maximum on the ground.

kg of objects; If the person can lift up to 40 kg of objects in an elevator that rises at a uniform acceleration, then the elevator ascent acceleration is .

m s2 Someone can lift up to 80 kg in an elevator that is descending with a uniform acceleration of acceleration, and up to a maximum of one on the ground.

kg of objects; If the person can lift up to 40 kg of objects in an elevator that rises at a uniform acceleration, then the elevator ascent acceleration is .

m/s2

When G becomes a uniform acceleration and falls, then he can lift 80*, and when it rises with a uniform acceleration 40*(g+a)=600, you will forget it, right?

When the elevator rises at a uniform acceleration, f-m3g=m3a3, and the substitution solution is a3=5m s2

According to the inscription, the man's strength n trace wide friend m(g-a) 80(

Therefore, the actual force of Zihuai, that is, the weight that can be lifted on the ground, is 584

At g = 10, the weight he can lift on the ground should be 60kg

When a person is in a lift that descends at a uniform acceleration, according to Newton's second law of lead:

mg-n=ma(1)

And Huai Qi because there is:

Understand the person's support for the object:

n=m(g-a)=80*(

then he can lift weights on the ground:

m = n zhiliang g = (600 10) kg = 60kg

Analyze the combined external force of the 80kg object in the descending elevator, due to the uniform acceleration of the fall, so the object out of the weightless state, that is, the object in the elevator and all the resultant force in the same direction of gravity is f=m(g-a)=80(, that is, Xiao Wang's arm can provide 600N, so the maximum weight that Xiao Wang can lift on the ground is 60kg- What a soft egg.

The gravitational force of an 80kg object on the ground is: 80*10 = 800nThe "gravity" of an object in the acceleration of the 80kg object in the elevator that accelerates and descends at a uniform acceleration is: 80*( 600n

A force of 600N can lift an object up to 60kg on the ground.

The force of 600N can only lift an object of 40kg in the elevator with uniform acceleration, so the acceleration is 600 40 - 10 = 5m s

Acceleration ascent Deceleration descent is overweight.

n=mg+mA=50 10+50 2=600nDeceleration and ascent Acceleration and descent is weightlessness.

n=mg-ma=50 10-50 2=400n, so choose ab

Hehe, if you don't know, please continue to ask.

a=v=

The moment the nut is dropped.

t1=v/a=

Because the lift accelerates upwards, the nut overweight acceleration is.

A cap = g + a =

The nut is lowered to the bottom with a time of t2

H = (1 2) A Cap * T 2

t2 = (2h a cap) = (2*.)

The displacement of the lift at this time.

s=(1 2)a(t1+t2)=(1 2)nut displacement.

s1=s-h=

Nut distance. s2=s+h=

With Leibniz's differential calculus,

Summary. Hello, paper questions can be photographed and sent to me.

Hello, paper questions can be photographed and sent to me.

The velocity of the ball decreases to zero at its highest point.

The ball rises in the same direction as the airplane, and the velocity is additive relative to the ground.

Hello, do you have any questions about this question?

A = when the nut falls t1 = v a = because the elevator accelerates upward, the overweight acceleration of the nut is a cap line grip finger = g + a = the nut is lowered to the bottom time t2h = (1 2) a cap * t 2t2 = (2h a cap) = 2 * the displacement of the elevator at this time s = (1 2) a (t1 + t2) = (1 2) the displacement of the nut.

s1=s-h=nut distance s2=s+h=a high school math problem.

A lift that starts to stand still accelerates the epithelium when the upshift speed is reached

Let the maximum lifting force of a person be f with an object as the object of study according to Newton's second law:

When the elevator to. When the acceleration of the uniform acceleration decreases, there is: m1

g-f=m1

a1 solves: f=600n

On the ground: the mass of the object that a person can lift is: m2fg60kg when the elevator rises at a uniform acceleration, f-m3

g=m3a3

Substitution solution: A3

5m s2 so the answer is: 60,5