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Scorpion4ik [409]
2 years ago
5

How much force would you need to accelerate a 4 kg object from a standstill to 5 m/s2?

Physics
1 answer:
djyliett [7]2 years ago
3 0

To solve this, we use Newtons 2nd law of motion which states that F = m⋅a . This means that a force F is required to make a body of mass m accelerate with an acceleration of a . So if a body of 4kg is to be accelerated by a force 17N then its acceleration can be found by putting into the formula F = m⋅a .

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1. Current is usually the flow of <u>electrons</u>.

Explanation:

The flow of charges through a wire or conductor is called electric current.

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Objects change motion due to forces. When they become unbalanced, their motion changes, while balanced forces cause constant mot
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Answer:

Balanced forces: When a number of forces acting on a body do not cause any change in its state of rest or of uniform motion along a straight line then the forces are said to be balanced forces. In other words, a body is said to be underbalanced forced when the resulting force acting on the body is zero.

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Example:

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2 years ago
What is the form of energy that batteries store energy as
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2 years ago
You pull on a spring whose spring constant is 22 N/m, and stretch it from its equilibrium length of 0.3 m to a length of 0.7 m.
Liono4ka [1.6K]

Answer:

W= 4.4 J

Explanation

Elastic potential energy theory

If we have a spring of constant K to which a force F that produces a Δx deformation is applied, we apply Hooke's law:

F=K*x  Formula (1): The force F applied to the spring is proportional to the deformation x of the spring.

As the force is variable to calculate the work we define an average force

F_{a} =\frac{F_{f}+F_{i}  }{2}  Formula (2)

Ff: final force

Fi: initial force

The work done on the spring is :

W = Fa*Δx

Fa : average force

Δx :  displacement

W = F_{a} (x_{f} -x_{i} )   :Formula (3)

x_{f} :  final deformation

x_{i}  :initial deformation

Problem development

We calculate Ff and Fi , applying formula (1) :

F_{f} = K*x_{f} =22\frac{N}{m} *0.7m =15.4N

F_{i} = K*x_{i} =22\frac{N}{m} *0.3m =6.6N

We calculate average force applying formula (2):

F_{a} =\frac{15.4N+6.2N}{2} = 11 N

We calculate the work done on the spring  applying formula (3) :         :

W= 11N*(0.7m-0.3m) = 11N*0.4m=4.4 N*m = 4.4 Joule = 4.4 J

Work done in stages

Work is the change of elastic potential energy (ΔEp)

W=ΔEp

ΔEp= Epf-Epi

Epf= final potential energy

Epi=initial potential energy

E_{pf} =\frac{1}{2} *k*x_{f}^{2}

E_{pi} =\frac{1}{2} *k*x_{i}^{2}

E_{pf} =\frac{1}{2} *22*0.7^{2} = 5.39 J

E_{pf} =\frac{1}{2} *22*0.3^{2} = 0.99 J

W=ΔEp=  5.39 J-0.99 J = 4.4J

:

4 0
2 years ago
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