A 20N crate starting at rest slides down a rough 5.0m ramp, inclined at 25 degress with the horizontal. 20j of energy is lost due to friction. what will be the speed of the crate at the bottom of the incline?
i need some help these one i don't know where to start
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PHYSICS - drwls, Tuesday, October 23, 2007 at 7:41pm
Is your @0N (the weight, M g) supposed to be 20N? I will assume so. If not, you provide the correct value.
The loss in potential energy is
M g * 5 sin 25 meters = 42.3 J
PE loss = friction + kinetic energy gain
42.3 = 20 + KE increase = (1/2) M V^2
(1/2) M V^2 = 22.3 J
M = 20 N/9.8 = 2.04 kg
V^2 = 2*22.3/2.04 m^2/s62
solve for V
ok the first time i did it i got 3.2m/s and the second time i did it i got 4.2m/s which is right?
Neither is right. You are making a computational mistake somewhere.
V = sqrt [2*22.3/2.04]
I get a number bigger than both of yours.
To find the speed of the crate at the bottom of the incline, we can use the principle of conservation of energy. Here's how to solve the problem step by step:
1. The first step is to calculate the loss in potential energy. The potential energy lost by the crate is equal to the work done by the force of gravity in moving the crate down the ramp. We can calculate this using the formula PE = mgh, where m is the mass of the crate, g is the acceleration due to gravity, and h is the vertical height of the ramp. In this case, h = 5 sin 25 (since the ramp is inclined at 25 degrees with the horizontal).
2. The next step is to find the amount of energy lost due to friction. In this case, it is given as 20 J.
3. Now, we can set up an equation for the conservation of energy:
Potential Energy loss = Frictional Energy loss + Kinetic Energy gain
So, PE = Friction + KE
42.3 J = 20 J + (1/2) m v^2
4. Rearrange the equation to solve for the speed of the crate, v:
42.3 J - 20 J = (1/2) m v^2
22.3 J = (1/2) m v^2
5. Substitute the values given in the problem:
m = 20 N / 9.8 m/s^2 (the weight divided by the acceleration due to gravity), which is approximately 2.04 kg.
6. Solve for v:
22.3 J = (1/2) * 2.04 kg * v^2
v^2 = (2 * 22.3 J) / 2.04 kg
v^2 ≈ 23.0 m^2/s^2
7. Finally, take the square root of both sides to find the magnitude of the speed, v:
v ≈ √23.0 m^2/s^2
v ≈ 4.8 m/s
Therefore, the speed of the crate at the bottom of the incline is approximately 4.8 m/s.
It seems that neither 3.2 m/s nor 4.2 m/s is the correct answer. The correct answer should be around 4.8 m/s according to the calculations above.