Answer: F = 131.7N
Explanation:
You are given the following parameters.
Mass M = 30 kg
Coefficient of static friction μ = 0.5
Ø = 30 degrees
When the person is trying to drag the box with force F, the static frictional force Fs will be acting in the opposite direction.
From the figure attached, resolve all forces into horizontal component and vertical component.
Horizontal component:
Fs - F cosØ = 0
Fs = F cosØ
F cosØ = μN ...... (1)
Vertical component:
N + F SinØ - mg = 0
N = Mg - F SinØ ..... (2)
Substitutes m, g and Ø into the equation 2
N = (30 × 9.8) - F × sin30
N = 294 - 0.5F
Substitute N and coefficient of friction into the equation (1)
F cos30 = 0.5 (294 - 0.5F)
Open the bracket
0.8660F = 147 - 0.25F
Collect the like terms
1.116025F = 147
F = 147/1.116025
F = 131.7 N
Therefore, the minimum force the person needs to have to move the box along the floor is 131.7 N
Answer:
Carbon 12
Explanation:
I don't 100% know what to put here, but...
When you remove the nucleus from an oxygen atom, almost everything of the base oxygen is essentially stripped away. Since almost everything is made of carbon, and Carbon 12 is one of the most common forms of Carbon, Carbon 12 would be what is left.
Answer: increase the amount of work that is done?
Explanation:
I’m not entirely sure to be honest but I’ll just use my logic here. Machines are made to increase efficiency and production. Machines are made to save time and energy as a pose to human work. So several machines can mass produce a lot of things and keep business booming. I’m not very good at explaining things but I hope this makes sense eventually.
Answer:
(a) v = 1.71 m/s
(b) μ = 0.005
Explanation:
(a)
Using the law of conservation of the momentum:
where,
m₁ = mass of person = 61.1 kg
m₂ = mass of sled = 16.1 kg
u₁ = initial speed of the person = 2.16 m/s
u₂ = initial speed of the sled = 0 m/s
v₁ = v₂ = v = final speeds of both the person and the sled = ?
Therefore,
<u>v = 1.71 m/s</u>
<u></u>
(b)
The kinetic energy lost by the sled must be equal to the frictional energy:
K.E = fd
where,
μ = coefficient of kinetic friction = ?
d = distance covered = 30 m
g = acceleration due to gravity = 9.81 m/s²
Therefore,
<u>μ = 0.005</u>
Elastic energy (E1) stored in the spring when it is stretched by x :
<span>=1/2kx^2=1.5J </span>
<span>when it is stretched 3 times more=>x becomes 3x
</span>so
<span>E2=1/2k(3x)^2 </span>
<span>=1/2k{9x^2} </span>
<span>=9{1/2kx^2} </span>
<span>={(E1}</span>
<span>=9{1.5} </span>
<span>=13.5J
hope it helps</span>