Anyone good at physics, that can help me? Will give 100 point!!! You don't need to show your work I just need the answer. Thank you.

Starting from rest, a 10 kg block slides 11.9 m down a frictionless ramp (inclined at
30◦ from the floor) to the bottom. The block then slides an additional 21.1 m along the floor before coming to stop. The acceleration of gravity is 9.8 m/s2.

A) Find the speed of the block at the bottom of the ramp.

B)Find the coefficient of kinetic friction between block and floor

C)Find the magnitude of the mechanical energy lost due to friction.

Hi there!

Part 1:

Assuming this is from the work-energy unit, we can use the work-energy theorem to solve.

Since the block starts from rest:

Initial energy = GPE = mgh

Final energy = KE = 1/2mv²

We are only given the diagonal distance of the ramp, so we must solve for its height using trigonometry.

sin(30) = O/H

Hsin(30) = 11.9sin(30) = 5.95 m

Now, solve by setting the GPE and KE equal:

mgh = 1/2mv²

gh = 1/2v²

v = √2gh

v = √2(9.8)(5.95) = 10.8 m/s

Part 2:

Now, we can use the equation for work:

W = ΔKE

The final velocity of the block after sliding is 0 m/s, so:

W = 1/2mv²

Recall the equation for work:

W = Fdcosθ

Since friction works AGAINST motion, cos(180) = -1.

Thus, the work done by friction is:

W = -Fd

Recall the equation for kinetic friction:

F = μmg

Thus:

0 = 1/2mv² - Fd

0 = 1/2mv² - μmg

μmgd = 1/2mv²

Cancel out the mass and rearrange to solve for the coefficient of friction:

μgd = 1/2v²

μd = 0.5v²/gd

μ = 0.5(10.8²)/(9.8)(21) = 0.283

Part 3:

The energy lost due to friction is equivalent to the WORK DONE by friction, which is equivalent to the initial kinetic energy of the block at the bottom of the ramp.

Thus:

W = 1/2mv² = Fd

W = 1/2(10)(10.8²) = 583.1 J

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