an athlete stretches a spring an extra 40.0 cm. how much energy has he transferred to the spring, if the spring constant is 5286.0 N/m?
Us= 1/2kx^2
=1/2(5286.0)(.40)=1057.2 J
Thanks!
Fixed Answer
Us= 1/2kx^2
=1/2(5286.0)(.40)^2=422.88 J
Well, look at Mr. Stretchy over here! If the athlete stretched the spring an extra 40.0 cm, we can calculate the energy transferred using the formula Us = 1/2 kx^2. So, plugging in the values, we get Us = 1/2 (5286.0) (.40) = 1057.2 J. So, the athlete has transferred a shocking 1057.2 Joules of energy to that spring! Time to bounce back with laughter!
To calculate the energy transferred to the spring, we can use the formula for potential energy stored in a spring:
Us = 1/2 * k * x^2
where Us is the potential energy stored, k is the spring constant, and x is the displacement of the spring.
Given that the spring constant is 5286.0 N/m and the athlete stretches the spring an extra 40.0 cm (or 0.40 m), we can substitute these values into the formula:
Us = 1/2 * 5286.0 * (0.40)^2 = 1057.2 J
Therefore, the athlete has transferred 1057.2 Joules of energy to the spring.
To calculate the amount of energy transferred to the spring, we can use the formula for potential energy stored in a spring, which is given by:
Us = (1/2)kx^2
Where:
Us is the potential energy stored in the spring
k is the spring constant
x is the displacement of the spring from its equilibrium position
In this case, the athlete stretches the spring an extra 40.0 cm, which is equal to 0.40 m. The spring constant is given as 5286.0 N/m.
Substituting these values into the formula:
Us = (1/2)(5286.0)(0.40)^2
= (1/2)(5286.0)(0.16)
= 1057.2 J
Therefore, the athlete has transferred 1057.2 Joules of energy to the spring.