Answer:
1.A
2.JU=mgh
U=(15)(9.8)(5)=735 J
3.Calculating energy Conservation of energy Types of energy Work-Energy Theorem Power Efficiency
4.The force exerted by the spring is given by Hooke's law:
F=-kxF=−kx
where
k is the spring constant, x is the distance when stretching the spring (and x is negative for compression),
and the elastic energy is
U=\dfrac{1}{2}kx^2U=21kx2
a) The spring constant is
k=\dfrac{F}{x}=\dfrac{375 \text{ N}}{0.05\text{ m}}=\bold{7500}\bold{\:N/m}k=xF=0.05 m375 N=7500N/m
b) When the spring is being stretched by 0.25 m, the force exerted by it is
F=kx=(7500\text{ N/m})(0.25\text{ m})=\bold{1875\:N}F=kx=(7500 N/m)(0.25 m)=1875N
c) The elastic energy is
U=\dfrac{1}{2}(7500\text{ N/m})(0.25\text{ m})^2=\bold{234.38\:J}U=21(7500 N/m)(0.25 m)2=234.38J
