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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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