A 21-N crate starting at rest slides down a rough 4-m long ramp, inclined at 25° with the horizontal. 15 J of energy is lost to friction. What will be the speed of the crate at the bottom of the incline?Select one:a.4.4 m/sb.1.4 m/sc.9.4 m/sd.3.1 m/se.19.1 m/s
Question
A 21-N crate starting at rest slides down a rough 4-m long ramp, inclined at 25° with the horizontal. 15 J of energy is lost to friction. What will be the speed of the crate at the bottom of the incline?Select one:a.4.4 m/sb.1.4 m/sc.9.4 m/sd.3.1 m/se.19.1 m/s
Solution
To solve this problem, we need to use the principle of conservation of energy. The total mechanical energy at the top of the ramp will be equal to the total mechanical energy at the bottom of the ramp, minus the energy lost due to friction.
Step 1: Calculate the gravitational potential energy (PE) at the top of the ramp. PE = mgh, where m is mass, g is the acceleration due to gravity, and h is the height.
First, we need to find the mass of the crate. We know that weight = mass * gravity. So, mass = weight / gravity = 21N / 9.8m/s^2 = 2.14 kg.
Next, we find the height of the ramp. h = length of ramp * sin(incline angle) = 4m * sin(25°) = 1.68m.
So, PE = 2.14kg * 9.8m/s^2 * 1.68m = 35.3 J.
Step 2: Calculate the kinetic energy (KE) at the bottom of the ramp. KE = PE - energy lost to friction = 35.3J - 15J = 20.3J.
Step 3: Solve for the speed at the bottom of the ramp. KE = 0.5 * m * v^2, where m is mass and v is speed. So, v = sqrt((2 * KE) / m) = sqrt((2 * 20.3J) / 2.14kg) = 4.4 m/s.
So, the speed of the crate at the bottom of the incline is 4.4 m/s. The answer is a. 4.4 m/s.
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