A 100kg crate is propelled uphill by a spring initially loaded with 400J of potential energy. the crate starts from rest. Point b is 0.2 m above point A. What is speed of the crate at point B?
initial energy = 400J
PE gained = mgh
total energy is conserved, so
KE = 400 - PE
1/2 mv^2 = KE, so solve for v
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To calculate the speed of the crate at point B, we need to consider the conservation of mechanical energy. Mechanical energy is the sum of kinetic energy (KE) and potential energy (PE).
At point A, the potential energy of the crate is 400J. At point B, the potential energy becomes zero because it is on the ground. Therefore, all the potential energy is converted into kinetic energy.
The equation representing mechanical energy conservation is:
KE_initial + PE_initial = KE_final + PE_final
Since the crate starts from rest at point A, the initial kinetic energy is zero:
0 + 400J = KE_final + 0
Simplifying:
400J = KE_final
To find the final kinetic energy (KE_final), we can use the equation:
KE = 0.5mv^2
Where:
KE = kinetic energy
m = mass of the crate
v = velocity (speed) of the crate
Since the mass of the crate is given as 100kg, we can substitute the values:
400J = 0.5 * 100kg * v^2
Simplifying further:
400J = 50kg * v^2
Dividing both sides of the equation by 50kg:
8J = v^2
Taking the square root of both sides:
√8J = v
Calculating the square root of 8J:
v ≈ 2.83 m/s
Therefore, the speed of the crate at point B is approximately 2.83 m/s.