The basketball is gaining kinetic and losing potential energy is the answer
40 N. If you’re using F=MA MxA would be 10x4 which would get you 40 N
Answer:
The percentage of the mechanical energy of the oscillator lost in each cycle is 6.72%
Explanation:
Mechanical energy (Potential energy, PE) of the oscillator is calculated as;
PE = ¹/₂KA²
During the first oscillation;
PE₁ = ¹/₂KA₁²
During the second oscillation;
A₂ = A₁ - 0.0342A₁ = 0.9658A₁
PE₂ = ¹/₂KA₂²
PE₂ = ¹/₂K (0.9658A₁)²
PE₂ = (0.9658²)¹/₂KA₁²
PE₂ = (0.9328)¹/₂KA₁²
PE₂ = 0.9328PE₁
Percentage of the mechanical energy of the oscillator lost in each cycle;

Therefore, the percentage of the mechanical energy of the oscillator lost in each cycle is 6.72%
At the highest point: kinetic energy is 0 due to the speed is 0
So the total mechanical energy is 20
Assume no frictions present, then the mechanical energy is conserved
So at the lowest point, kinetic energy = mechanical energy - potential energy
Answer will be 20 - 0.5 = 19.5 J