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Aloiza [94]
2 years ago
11

What is the change in thermal energy E if the coefficent of kinetic friction between the box and floor is .4 , the distance the

box moves is 17m and the force applied is 38 N?
Physics
1 answer:
Jet001 [13]2 years ago
7 0

This question can be solved using the concept of friction energy.

The thermal energy change is b "258.4 J".

The change in thermal energy will be equal to the friction energy produced during the motion of the box.

Change\ In\ Thermal\ Energy = E = Friction\ Energy\\\\E = \mu fd

where,

μ = coefficient of kinetic friction = 0.4

f = force applied = 38 N

d = distance traveled by the box = 17 m

Therefore,

E = (0.4)(38\ N)(17\ m)

<u>E = 258.4 J</u>

Learn more about friction energy here:

brainly.com/question/1343045?referrer=searchResults

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3 years ago
A block of wood is 4 cm wide, 5 cm long, and 10 cm high. It weighs 100 grams. Calculate its volume. Calculate its density. Will
77julia77 [94]

Answer:

V=200cm^3\\\\\rho =0.500g/cm^3

It will float.

Explanation:

Hello.

In this case, given the width, length and height, we can compute the volume as follows:

V=W*L*H\\\\V=4cm*5cm*10cm\\\\V=200cm^3

Moreover, since the density is computed via the division of the mass by the volume:

\rho =\frac{m}{V}

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Read 2 more answers
A block rests on a frictionless horizontal surface and is attached to a spring. When set into simple harmonic motion, the block
MariettaO [177]

Answer:

v₀ = 0.5058 m/s

Explanation:

From the question, for the block to hit the bottle, the elastic potential energy of the spring at the bottle (x = 0.08 m) should be equal to the sum of the elastic potential energy of the spring at x = 0.05 m and the kinetic energy of block at x = 0.05 m

Now, the potential energy of the block at x = 0.08 m is ½kx²

where;

k is the spring constant given by; k = ω²m

ω is the angular velocity of the oscillation

m is the mass of the block.

Thus, potential energy of the spring at the bottle(x = 0.08 m) is;

U = ½ω²m(0.08m)²

Also, potential energy of the spring at the bottle(x = 0.05 m) is;

U = ½ω²m(0.05m)²

and the kinetic energy of the block at x = 0.05 m is;

K = ½mv₀²

Thus;

½ω²m(0.08)² = ½ω²m(0.05)² + ½mv₀²

Inspecting this, ½m will cancel out to give;

ω²(0.08)² = ω²(0.05)² + v₀²

Making v₀ the subject, we have;

v₀ = ω√((0.08)² - (0.05)²)

So,

v₀ = 8.1√((0.08)² - (0.05)²)

v₀ = 0.5058 m/s

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