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A block of mass 3.5 kg slides down a frictionless inclined plane of length 6.4 m that makes an angle of 30° with the horizontal. If the block is released from rest at the top of the incline, what is its speed at the bottom?(A) 5.0 $\mathrm{m} / \mathrm{s}$(B) 6.4 $\mathrm{m} / \mathrm{s}$(C) 8.0 $\mathrm{m} / \mathrm{s}$(D) 10 $\mathrm{m} / \mathrm{s}$

Physics 101 Mechanics

Chapter 5

Work, Energy, and Power

Work

Kinetic Energy

Energy Conservation

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this question. We have blocked that sliding down, uh, on a fiction is inclined Plane. Yeah. So? Okay. And the distance distance between the starting point and the endpoint is, uh, 6.4 m. Hey. And then the anger here is 30 degrees. I have a block. Okay, that slides down on a perfectionist and climb pain. And then this long is 3.5 kg. We want to find, uh, speed off the block at the bottom when it is released from rest. Okay. At the top. Yeah. So to solve this problem, what he even be using, um, the conservation of energy. Okay, so thesis is will be the initial, and then here would be the final. Okay, then we have k I plus new. I goes to k f plus U f K initially the, um, kinetic energy zero, because the block is initially at rest, and then the u i S m g the h. So the H is is systems the age, then care is happening. The square. Okay. The final potential energy is zero. Okay, so, um, so from this triangle that we have over here, Okay, we have here 30 degrees here is the goes to 6.4 m. Okay, so this is the package. So we have data age over the equals to sign 30 degrees. So the age because to be sign degrees. Okay, so we have, uh, half mv squared. It goes to m g E side 30 degrees. Okay. Can cancel out. So obvious to g d sign the degrees square roots. Yeah. Two times. 10 times 6.47 30 degrees is 0.5. Yeah, he saw the two and 0.5 becomes one. And so you have square roots up 64 and you get 8.0. Does hurt second. So the answer. Yes. Um, see. Okay, so this is the answer. America answer. And then this is the, uh, have a bit answer. Okay? And that's all for this question.

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