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Sergio039 [100]

2 years ago

7

A 5.0 kg box slides down a 4.0 m long ramp that makes a 25 angle with the ground. If the coefficient of kinetic friction is 0.65

, how much thermal energy was produced?

1 answer:

Sedaia [141]2 years ago

4 0

The thermal energy was produced is 116J

<h3>What is the thermal energy produced?</h3>

Now we know that the frictional force produces the energy that is lost as heat as the body slides down the incline. The magnitude of the frictional force is obtained from;

Ff= μmgcosθ

Ff = 0.65 * 5.0 kg * 9.8 m/s^2 * cos 25

Ff = 29 N

Hence, the thermal energy is;

29 N * 4.0 m = 116J

Learn more about frictional force:brainly.com/question/1714663

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In the physics lab, a block of mass M slides down a frictionless incline from a height of 35cm. At the bottom of the incline it

bogdanovich [222]

Solution :

Given :

M = 0.35 kg

$m=\frac{M}{2}=0.175 \ kg$

Total mechanical energy = constant

or $K.E._{top}+P.E._{top} = K.E._{bottom}+P.E._{bottom}$

But $K.E._{top} = 0$ and $P.E._{bottom} = 0$

Therefore, potential energy at the top = kinetic energy at the bottom

$\Rightarrow mgh = \frac{1}{2}mv^2$

$\Rightarrow v = \sqrt{2gh}$

$=\sqrt{2 \times 9.8 \times 0.35}$ (h = 35 cm = 0.35 m)

= 2.62 m/s

It is the velocity of M just before collision of 'm' at the bottom.

We know that in elastic collision velocity after collision is given by :

$v_1=\frac{m_1-m_2}{m_1+m_2}v_1+ \frac{2m_2v_2}{m_1+m_2}$

here, $m_1=M, m_2 = m, v_1 = 2.62 m/s, v_2 = 0$

∴ $$v_1=\frac{0.35-0.175}{0.5250}+\frac{2 \times 0.175 \times 0}{0.525}$

$=\frac{0.175}{0.525}+0$

= 0.33 m/s

Therefore, velocity after the collision of mass M = 0.33 m/s

3 0

3 years ago

You hang a heavy ball with a mass of 10 kg from a gold wire 2.6 m long that is 1.6 mm in diameter. You measure the stretch of th

PolarNik [594]

<u>Answer:</u> The Young's modulus for the wire is $6.378\times 10^{10}N/m^2$

<u>Explanation:</u>

Young's Modulus is defined as the ratio of stress acting on a substance to the amount of strain produced.

The equation representing Young's Modulus is:

$Y=\frac{F/A}{\Delta l/l}=\frac{Fl}{A\Delta l}$

where,

Y = Young's Modulus

F = force exerted by the weight = $m\times g$

m = mass of the ball = 10 kg

g = acceleration due to gravity = $9.81m/s^2$

l = length of wire = 2.6 m

A = area of cross section = $\pi r^2$

r = radius of the wire = $\frac{d}{2}=\frac{1.6mm}{2}=0.8mm=8\times 10^{-4}m$ (Conversion factor: 1 m = 1000 mm)

$\Delta l$ = change in length = 1.99 mm = $1.99\times 10^{-3}m$

Putting values in above equation, we get:

$Y=\frac{10\times 9.81\times 2.6}{(3.14\times (8\times 10^{-4})^2)\times 1.99\times 10^{-3}}\\\\Y=6.378\times 10^{10}N/m^2$

Hence, the Young's modulus for the wire is $6.378\times 10^{10}N/m^2$

3 0

4 years ago