Pete Zaria applies a 4.0-N force to a 1.0-kg mug of root beer to accelerate it over a distance of 1.0-meter along the countertop. Determine the work done by Pete on the mug and the mug's final kinetic energy and final velocity. PSYW

Respuesta :

Answer:

Work done = 4 J

Final Kinetic Energy = 4 J

Explanation:

From the question,

Work done = force × distance

W = F×d................... Equation 1

Given: F = 4.0 N, d = 1.0 meter.

Substitute these values into equation 1

W = 4×1

W = 4 Joules.

Also,

Kenetic Energy = 1/2 × mass × velocity×velocity.

K.E = 1/2(mv²)........... (2).

But,

F = ma

Where F = 4 N, m = 1 kg.

a = 4/1 = 4 m/s².

Using,

v² = u² + 2as............. Equation 3

Where u = 0 m/s, a = 4 m/s², s = 1 m.

Substitute into equation 3

v² = 0² +2×4×1

v² = 8

v = √8 m/s.

Substitutting into equation 2

K.E = 1/2(1)(√8)²

K.E = 1/2(8)

K.E = 4 J.

Hence the work done and the Final Kinetic energy are thesame = 4 J

The work done by the Pete is 4.0 J. And its final velocity and kinetic energy are 2.82 m/s and 3.97 J respectively.

Given data:

The force applied by the Pete Zaria is, F = 4.0 N.

The mass of mud is, m = 1.0 kg.

The distance of the countertop is, s = 1.0 m.

The work by the Pete on the mug is equal to the product of applied force and the distance covered by the mug along the rooftop. Therefore,

W = F × s

W = 4.0 × 1.0

W = 4.0 J

Now, apply the third kinematic equation of motion to obtain the final velocity of the mug as,

[tex]v^{2} = u^{2}+2as[/tex]

Here, a is the magnitude of acceleration and its value is,

F = ma

4.0 = (1.0)a

a = 4.0 m/s²

Now,

[tex]v=\sqrt{0^{2}+(2 \times 4 \times 1)}\\\\v = 2.82 \;\rm m/s[/tex]

Now. the kinetic energy of the mug is calculated as,

[tex]KE = \dfrac{1}{2}mv^{2}\\\\KE = \dfrac{1}{2} \times 1.0 \times 2.82^{2}\\\\KE = 3.97 \;\rm J[/tex]

Thus, we can conclude that the work done by the Pete is 4.0 J. And its final velocity and kinetic energy are 2.82 m/s and 3.97 J respectively.

Learn more about the kinetic energy here:

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