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4 years ago

When forces are balanced, what affect does that have on the motion of an object?

Physics

2 answers:

salantis [7]4 years ago

7 0

Balanced forces cause no change in motion, even if the object is moving.

ElenaW [278]4 years ago

5 0

According to Newton’s 1st law, all objects remain stable unless acted upon by an unbalanced force, so this means that since the forces are balanced, the motion would remain at rest. If there was a greater force on one side of the object, then it would be pushed to the other side (ex. Pushing a heavy box from the right, makes it go left). I hope that helps...

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What are 3 types of contact forces

Svetach [21]

There are different types of contact forces like normal Force, spring force, applied force and tension force.

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3 years ago

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What is the smallest particle that relains<br> the properties of an element

VLD [36.1K]

Answer:

atom

Explanation:

Smallest particle of an element that identifies that element is an atom.

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2 years ago

A 36.0 kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of 130

konstantin123 [22]

Answer:

(a) W = 650J

(b) Wf = 529.2J

(d) W = 0J

(e) ΔK = 120.8J

(f) v2 = 2.58 m/s

Explanation:

(a) In order to find the work done by the applied force you use the following formula:

$W=Fd$ (1)

F: applied force = 130N

d: distance = 5.0m

$W=(130N)(5.0m)=650J$

The work done by the applied force is 650J

(b) The increase in the internal energy of the box-floor system is given by the work done of the friction force, which is calculated as follow:

$W_f=F_fd=\mu Mgd$ (2)

μ: coefficient of friction = 0.300

M: mass of the box = 36.0kg

g: gravitational constant = 9.8 m/s^2

$W_f=(0.300)(36.0kg)(9.8m/s^2)(5.0m)=529.2J$

The increase in the internal energy is 529.2J

(c) The normal force does not make work on the box because the normal force is perpendicular to the motion of the box.

$W = 0J$

(d) The same for the work done by the normal force. The work done by the gravitational force is zero because the motion of the box is perpendicular o the direction of the gravitational force.

(e) The change in the kinetic energy is given by the net work on the box. You use the following formula:

$\Delta K=W_T$ (3)

You calculate the total work:

$W_T=Fd-F_fd=(F-F_f)d$ (4)

F: applied force = 130N

Ff: friction force

d: distance = 5.00m

The friction force is:

$F_f=(0.300)(36.0kg)(9.8m/s^2)=105.84N$

Next, you replace the values of all parameters in the equation (4):

$W_T=(130N-105.84N)(5.00m)=120.80J$

$\Delta K=120.80J$

The change in the kinetic energy of the box is 120.8J

(e) The final speed of the box is calculated by using the equation (3):

$W_T=\frac{1}{2}M(v_2^2-v_1^2)$ (5)

v2: final speed of the box

v1: initial speed of the box = 0 m/s

You solve the equation (5) for v2:

$v_2 = \sqrt{\frac{2W_T}{M}}=\sqrt{\frac{2(120.8J)}{36.0kg}}=2.58\frac{m}{s}$

The final speed of the box is 2.58m/s

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3 years ago

What is the definition of elastic limit in hookes law?

notsponge [240]

Within the elastic limit of a solid material, the deformation (strain) produced by a force (stress) of any kind is proportional to the force. If the elastic limit is not exceeded, the material returns to it original shape and size after the force is removed, other it remains deformed or stretched.

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3 years ago

2 kg<br> Use 10 m/s2 for g

Iteru [2.4K]

Answer:

See below

Explanation:

At point A the PE = mgh = 2 * 10 * 1 = 20 J

at point B, all of the PE , 20 J , is converted to Kinetic Energy

KE = 1/2 m v^2

20 = 1/2 (2)(v^2 )

20 = v^2 v = sqrt 20 = 4.47 m/s

for the friction part

vf = vo t + 1/2 a t^2 vf = final velocity = 0 (stopped)

vo = original velocity = 4.47 m/s

a = -1 m/s^2

0 = 4.47 t + 1/2 (-1) t^2

- .5t^2 + 4.47 t = 0

t ( -.5t+ 4.47) = 0 shows t = 4.47/.5 = 8.9 seconds

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2 years ago