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Gemiola [76]
3 years ago
10

A box of mass 26 kg is initially at rest on a flat floor. The coefficient of kinetic friction between the box and the floor is 0

.60. A woman pushes horizontally against the box with a force of 505 N until the box attains a speed of 2 m/s. What is the change in kinetic energy of the box?
Physics
1 answer:
Kazeer [188]3 years ago
3 0

Answer:

\Delta K = 52J

Explanation:

The change in kinetic energy will be simply the difference between the final and initial kinetic energies: \Delta K=K_f-K_i

We know that the formula for the kinetic energy for an object is:

K=\frac{mv^2}{2}

where <em>m </em>is the mass of the object and <em>v</em> its velocity.

For our case then we have:

\Delta K = K_f-K_i=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}=\frac{m(v_f^2-v_i^2)}{2}

Which for our values is:

\Delta K = \frac{m(v_f^2-v_i^2)}{2} = \frac{(26Kg)((2m/s)^2-(0m/s)^2)}{2} = 52J

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A uniform cylinder of radius 25 cm and mass 27 kg is mounted so as to rotate freely about a horizontal axis that is parallel to
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Answer:

Explanation:

                                                     STEP 1

<u>Given</u>

Radius of cylinder = r = 25cm, 2.5m

mass = 27kg

cylinder is mounted so as to rotate freely about a horizontal axis that is parallel to and 60cm to the central logitudinal axis of the cylinder

height = 0.6m

<u>part 1</u>

The cylinder is mounted so as to rotate freely about a horizontal axis tha is paralle to 60cm from the central longitudinal axis of then cylinder. The rotational inertia of the cylinder about the axis of rotation is given by

<em>I = Icm + mh²</em>

<em>∴ I = 1/2mr² + mh² = 1/2x27x (0.5)² + 20  x  (0.6)²</em>

<em>I=13.09kg.m²</em>

where

<em>I</em>cm is the rotational inertia of the cylinder about its central axis

m is the mass of the cylinder

h is the distance between the axis of the rotation and the central axis of the cylinder

r is the radius of the cylinder

<em>                                        </em><em> I=13.09kg.m²</em>

<em>part2</em>

<em>from the conservation of the total mechanical energy of the meter stick, the change in gravitational potential energyof the meter stick plus the change in kinetic energy must be zero</em>

<em>Δk + Δu = 0</em>

<em>1/2 </em>I(w²-w²) = Ui-Uf

1/2 x 13.09w² = mgh

∴w=√20 x 9.8 x 0.6/(1/2 x 13.09) =117.6/6.5

w=18.09rad/s

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