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andrew-mc [135]
3 years ago
6

A 15.0 kg crate, initially at rest, slides down a ramp 2.0 m long and inclined at an angle of 20.0° with the horizontal. Using t

he work-kinetic energy theorem and disregarding friction, find the
velocity of the crate at the bottom of the ramp. (g = 9.81 m/s?)
Physics
1 answer:
Elis [28]3 years ago
3 0

The component of the crate's weight that is parallel to the ramp is the only force that acts in the direction of the crate's displacement. This component has a magnitude of

<em>F</em> = <em>mg</em> sin(20.0°) = (15.0 kg) (9.81 m/s^2) sin(20.0°) ≈ 50.3 N

Then the work done by this force on the crate as it slides down the ramp is

<em>W</em> = <em>F d</em> = (50.3 N) (2.0 m) ≈ 101 J

The work-energy theorem says that the total work done on the crate is equal to the change in its kinetic energy. Since it starts at rest, its initial kinetic energy is 0, so

<em>W</em> = <em>K</em> = 1/2 <em>mv</em> ^2

Solve for <em>v</em> :

<em>v</em> = √(2<em>W</em>/<em>m</em>) = √(2 (101 J) / (2.0 m)) ≈ 10.0 m/s

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The body weighing 2 kg moves through the horizontal surface and crosses the path x = 75 cm The coefficient of friction of the bo
Taya2010 [7]

The kinetic energy of the body in definitive position is 4.24 J.

Explanation:

As per the work energy theorem, the work done on any system or object to move it from one position to another is equal to the change in kinetic energy of the object. In this case, the body weighing 2 kg is moved over an horizontal surface for a distance of 75 cm. As there will be frictional force acting on the body while moving over the surface. This frictional force multiplied by the distance the object is moved will give the work done on the body.

Frictional force = Coeffficent of friction × Normal force.

As the weight of the body is 2 kg, the normal force acting on it will be mass multiplied with acceleration due to gravity.

Frictional force = - 0.8×9.8 × 2 =-15.68 N

So the work done will be the product of frictional force with the displacement of 75 cm or 0.75 m.

Work done =  Frictional force × Displacement

Work done = -15.68×0.75 = -11.76 J.

So the work is done by the object.

If the kinetic energy of the body at starting is 16 J, then the kinetic energy of the body at definitive position will be obtained as below.

Work done = change in kinetic energy

-11.76 J = Final kinetic energy-16 J

Final Kinetic energy = - 11.76+16

Final kinetic energy = 4.24 J

Thus, the kinetic energy of the body in definitive position is 4.24 J.

3 0
2 years ago
Please help with this!!!!!
34kurt
(1.00 atm) (0.1156 L) = (n) (0.08206 L atm / mol K) (273 K)  I hoped that helped
4 0
3 years ago
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