A retractable barrier uses a spring. During a test, a force of 5.0 N holds the spring at an extension of 0.020 m. In use, the same spring is stretched to an extension of 0.060 m. The force-extension relationship is linear.

Use $F = k x$ and $E=12kx2\displaystyle E=\frac{1}{2}kx^2$ .

What is the correct calculation for the elastic energy stored in the spring when the barrier is in use?

$12×(5.00.020)×(0.060)2\dfrac{1}{2}\times\left(\dfrac{5.0}{0.020}\right)\times(0.060)^2$

$12×(0.0205.0)×(0.060)2\dfrac{1}{2}\times\left(\dfrac{0.020}{5.0}\right)\times(0.060)^2$

$12×(5.00.060)×(0.060)2\dfrac{1}{2}\times\left(\dfrac{5.0}{0.060}\right)\times(0.060)^2$

$12×(5.00.020)×(0.020)2\dfrac{1}{2}\times\left(\dfrac{5.0}{0.020}\right)\times(0.020)^2$

Use $F = k x$ and $E=12kx2\displaystyle E=\frac{1}{2}kx^2$ .

What is the correct calculation for the elastic energy stored in the spring when the barrier is in use?

$12×(5.00.020)×(0.060)2\dfrac{1}{2}\times\left(\dfrac{5.0}{0.020}\right)\times(0.060)^2$

$12×(0.0205.0)×(0.060)2\dfrac{1}{2}\times\left(\dfrac{0.020}{5.0}\right)\times(0.060)^2$

$12×(5.00.060)×(0.060)2\dfrac{1}{2}\times\left(\dfrac{5.0}{0.060}\right)\times(0.060)^2$

$12×(5.00.020)×(0.020)2\dfrac{1}{2}\times\left(\dfrac{5.0}{0.020}\right)\times(0.020)^2$

15.1 Forces and springs

Question 2

Markscheme

15.1 Forces and springs

Question 2

Markscheme

15.1 Forces and springs

Question 2

15.1 Forces and springs Questions

15.1 Forces and springs

Question 2

15.1 Forces and springs Questions