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IRISSAK [1]
3 years ago
15

A forklift raises a 1020 N crate 3.50 m up to a shelf l. How much work is done by the forklift on the crate? The forklift does J

work on the crate.
Physics
2 answers:
Rom4ik [11]3 years ago
7 0

A forklift raises a 1,020 N crate 3.50 m up to a shelf. How much work is done by the forklift on the crate?

The forklift does  

⇒ 3,570 J of work on the crate. the answer is 3570

lord [1]3 years ago
5 0

Answer : Work done on the crate is 3570 J

Explanation :

Force acting on forklift due to its mass is 1020 N

Distance covered in lifting it, d = 3.5 m

Mathematically, the work done is defined as :

W=F\times d

So, W=1020\ N\times 3.5\ m

W=3570\ J

The forklift does 3570 J work on the crate.      

Hence, this is the required solution.                      

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Part A A uniform solid disk of radius 1.60 m and mass 2.30 kg rolls without slipping to the bottom of an inclined plane. If the
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Answer:

Height will be 3.8971 m        

Explanation:

We have given that radius of the solid r = 1.60 m

Mass of the solid disk m = 2.30 kg

Angular velocity \omega =4.46rad/sec

Moment of inertia is given by I=\frac{1}{2}mr^2

Transnational Kinetic energy is given by KE=\frac{1}{2}mv^2 as we know that v = v=\omega r

So KE=\frac{1}{2}m(\omega r)^2

Rotational kinetic energy is given by KE_{ROTATIONAL}=\frac{1}{2}I\omega ^2=\frac{1}{2}\left ( \frac{1}{2}mr^2 \right )\omega ^2=\frac{1}{4}m(r\omega )^2

Potential energy is given by mgh

According to energy conservation

mgh=\frac{1}{2}m(\omega r)^2+\frac{1}{4}m(\omega r)^2

h=\frac{3r^2\omega ^2}{4g}=\frac{3\times 1.60^2\times 4.46^2}{4\times 9.8}=3.8971m

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3 years ago
A 2.45-kg frictionless block is attached to an ideal spring with force constant 355 N/m. Initially the spring is neither stretch
ANTONII [103]

Answer:

A.    A = 0.913 m

B.    amax = 132.24m/s^2

C.    Fmax = 324.01N

Explanation:

When the block is moving at the equilibrium point , its velocity is maximum.

A. To find the amplitude of the motion you use the following formula for the maximum velocity:

v_{max}=A\omega          (1)

vmax = maximum velocity = 11.0 m/s

A: amplitude of the motion = ?

w: angular frequency = ?

Then, you have to calculate the angular frequency of the motion, by using the following formula:

\omega=\sqrt{\frac{k}{m}}           (2)

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\omega = \sqrt{\frac{355N/m}{2.45kg}}=12.03\frac{rad}{s}

Next, you solve the equation (1) for A and replace the values of vmax and w:

A=\frac{v}{\omega}=\frac{11.0m/s}{12.03rad/s}=0.913m

The amplitude of the motion is 0.913m

B. The maximum acceleration of the block is given by:

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The maximum acceleration is 132.24 m/s^2

C. The maximum force is calculated by using the second Newton law and the maximum acceleration:

F_{max}=ma_{max}=(2.45kg)(132.24m/s^2)=324.01N

It is also possible to calculate the maximum force by using:

Fmax = k*A = (355N/m)(0.913m) = 324.01N

The maximum force exertedbu the spring on the object is 324.01 N

4 0
2 years ago
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