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1 year ago

an object has a mass of 50kg, a final height of 20m and an initial height of 8m. what is the amount of work done

Physics

1 answer:

Andrei [34K]1 year ago

7 0

amount of work done is 5880 J

Given:

mass of object = 50kg

Final height = 20m

initial height = 8m

To Find:

amount of work done

Solution:

work is done when a force acts upon an object to cause a displacement. You can calculate the energy transferred, or work done, by multiplying the force by the distance moved in the direction of the force.

The work done by gravity is given by the formula,

W = mgh

W = 50 x 9.8 x ( 20-8)

= 5880 J

So the work done is 5880 J

Learn more about Work done here:

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Four blocks of weights are required using which any body whose weight is between 1kg and 40 kg can be weighed. Find the four wei

Vikentia [17]

Answer:

The weights are 1 kg, 3kg, 9kg and 27kg.

Explanation:

The weights are 1 kg, 3kg, 9kg and 27kg.

1+3+9+27= 40

27+9+3= 39

27+9+3-1=38

27+9+1=37

27+9=36

27+9-1=35

27+9+1-3=34

27+9-3=33

27+9-3-1=32

27+3+1=31

27+3=30

27+3-1=29

27+1=28

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27-3=24

27-3-1=23

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Like this all the weights from 1 to 40 kg can be made using 1,3,9 and 27 kg.

6 0

2 years ago

A factory worker pushes a crate of mass 31.0 kg a distance of 4.35 m along a level floor at constant velocity by pushing horizon

Debora [2.8K]

Answer:

a. 79.1 N

b. 344 J

c. 344 J

d. 0 J

e. 0 J

Explanation:

a. Since the crate has a constant velocity, its net force must be 0 according to Newton's 1st law. The push force $F_p$ by the worker must be equal to the friction force $F_f$ on the crate, which is the product of friction coefficient μ and normal force N:

Let g = 9.81 m/s2

$F_p = F_f = \mu N = \mu mg = 0.26 * 31 * 9.81 = 79.1 N$

b. The work is done on the crate by this force is the product of its force $F_p$ and the distance traveled s = 4.35

$W_p = F_ps = 79.1*4.35 = 344 J$

c. The work is done on the crate by friction force is also the product of friction force and the distance traveled s = 4.35

$W_f = F_fs = -79.1*4.35 = -344 J$

This work is negative because the friction vector is in the opposite direction with the distance vector

d. As both the normal force and gravity are perpendicular to the distance vector, the work done by those forces is 0. In other words, these forces do not make any work.

e. The total work done on the crate would be sum of the work done by the pushing force and the work done by friction

$W_p + W_f = 344 - 344 = 0 J$