A block of ice of mass 30 kg is at rest atop an inclined plane of vertical height
2 m. It is released and slides down the inclined plane. What is its speed at the bottom of the inclined plane?
speedbottom:
KE at bottom=PE at top
1/2 mv^2=mgh
v^2=2gh=2*9.8*2
speedbottom=sqrt(4*9.8) m/s
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a
To find the speed of the block of ice at the bottom of the inclined plane, we can use the principle of conservation of energy.
1. First, we need to find the potential energy of the block of ice at the top of the inclined plane.
Potential energy (PE) = mass * gravity * height
PE = 30 kg * 9.8 m/s^2 * 2 m
PE = 588 J
2. Next, we need to find the kinetic energy of the block of ice at the bottom of the inclined plane.
Since the block of ice is at rest at the top of the inclined plane, its initial kinetic energy is zero.
3. According to the conservation of energy, the total mechanical energy (potential energy + kinetic energy) is conserved.
Total Mechanical Energy at the top = Total Mechanical Energy at the bottom
Potential Energy at the top = Kinetic Energy at the bottom
PE = KE
588 J = (1/2) * mass * velocity^2
4. Rearranging the equation, we can solve for the velocity (v):
velocity^2 = (2 * PE) / mass
velocity^2 = (2 * 588 J) / 30 kg
velocity^2 = 39.2 m^2/s^2
5. Taking the square root of both sides, we can find the velocity (v):
velocity = √(39.2 m^2/s^2)
velocity ≈ 6.26 m/s
Therefore, the speed of the block of ice at the bottom of the inclined plane is approximately 6.26 m/s.
To find the speed of the block of ice at the bottom of the inclined plane, we can use the principle of conservation of energy.
The potential energy of the block at the top of the inclined plane (when it is at rest) will be converted into kinetic energy at the bottom of the inclined plane.
First, let's calculate the potential energy of the block at the top of the incline. The potential energy (PE) is given by the formula:
PE = m * g * h
where m is the mass of the block (30 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the vertical height of the inclined plane (2 m).
Substituting the values into the formula, we have:
PE = 30 kg * 9.8 m/s^2 * 2 m
= 588 Joules
Since the potential energy is converted into kinetic energy at the bottom of the inclined plane, we can equate the potential energy to the kinetic energy (KE) using the equation:
KE = 0.5 * m * v^2
where v is the velocity or speed of the block at the bottom of the inclined plane.
Substituting the values, we have:
588 Joules = 0.5 * 30 kg * v^2
Simplifying the equation, we get:
v^2 = (2 * 588 Joules) / 30 kg
v^2 = 39.2 Joules / kg
Taking the square root of both sides, we find:
v = √(v^2)
v = √(39.2 Joules / kg)
v ≈ 6.26 m/s
Therefore, the speed of the block of ice at the bottom of the inclined plane is approximately 6.26 m/s.