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fomenos
2 years ago
10

While dragging a crate a workman exerts a force of 628 N. Later, the mass of the crate is increased by a factor of 3.8. If the w

orkman exerts the same force, how does the new acceleration compare to the old acceleration?
Physics
2 answers:
Delvig [45]2 years ago
6 0
Force applied = F = 628 N 
<span>Acceleration = a m/s² </span>
<span>Newton's 2nd law of motion : F = Ma </span>
<span> a = F/M -------- (1) </span>
<span>New mass of the crate = M1 = 3.8M kg </span>
<span>New acceleration = a1 = F/M1 = F/(3.8 M) ----- (2) </span>
<span>a1/a = {F/(3.8M)}/(F/M) = 1/3.8 = 10/38 = 5/19 ------- Answer</span>
Alja [10]2 years ago
5 0

Answer:

The acceleration would decrease by a factor of 3.8.

Explanation:

Force = mass x acceleration

F = ma

If the mass is increased by a factor of 3.8. New mass = 3.8 m

If the same force is exerted, then acceleration is inversely proportional to mass and it changes by:

\frac {3.8 m}{m} = \frac{a} {a'}\\  \Rightarrow a' = \frac {a} {3.8}

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A box is at rest on a ramp at an incline of 22°. The normal force on the box is 538 N.
fomenos

Answer: 580 N

Refer to attached figure.

The angle of inclination is 22 degrees

weight (gravitational force) acts downwards.

Normal force is a contact force which acts perpendicular to the point of contact.

The horizontal component (mg cos 22 ) balances the normal force and the vertical component balances the frictional force.

Gravitational force on an object = mg

The normal force N= mg cos 22

\Rightarrow mg =\frac{N}{cos22}=\frac{538 N}{0.927}=580 N





8 0
2 years ago
Read 2 more answers
The box leaves position x=0x=0 with speed v0v0. The box is slowed by a constant frictional force until it comes to rest at posit
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Answer:

fr = ½ m v₀²/x

Explanation:

This exercise the body must be on a ramp so that a component of the weight is counteracted by the friction force.

The best way to solve this exercise is to use the energy work theorem

            W = ΔK

Where work is defined as the product of force by distance

           W = fr x cos 180

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          Δk =K_{f} –K₀

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We substitute

         - fr x = - ½ m v₀²      

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8 0
2 years ago
Military specifications often call for electronic devices to be able to withstand accelerations of 10 g. to make sure that their
trapecia [35]
The solution for this problem is:
(10 x 9.8) = 98.1 m/sec^2 acceleration. Time, to travel 9.4cm or (.094m.), at acceleration of 98m/sec^2

= sqrt(2d/a), = sqrt (98.1 m/sec^2/0.094m) = 32.3050619 sec per cycle

Frequency = (w/2pi), = 32.3050619/2pi
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"an object keeps its state of rest or of uniform motion in a straight line unless acted upon by an external net force different from zero"

This means that if there are no forces acting on an object, the object stays at rest (if it was not moving previously) or it continues moving with same velocity (if it was already moving) in a straight line.

In this problem, the Earth is initially moving around the Sun, with a certain tangential velocity v. When the Sun disappears, the force of gravity that was keeping the Earth in circular motion disappears too: therefore, there are no more forces acting on the Earth, and so by the 1st law of Newton, the Earth will continue moving with same velocity v in a straight line.

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