9702_s22_qp_12 Question 13

A diving board of length 5.0 m is hinged at one end and supported 2.0 m from this end by a spring of spring constant 10 kN m^–1.  A child of mass 40 kg stands at the far end of the board. 

What is the extra compression of the spring caused by the child standing on the end of the board? 

A 1.0 cm 

B 1.6 cm 

C 9.8 cm 

D 16 cm

ANSWER: C

The Physics Behind

• To determine the compression x of the spring, Hooke's Law will be used.

x = F / k

where F is the force experienced by spring and k is the spring constant.

• The force F can be calculated using the second condition of equilibrium.

(weight) (5.0 m) = (F) (2.0 m)

F' = ((40 kg) (9.81 N kg-1) (5.0 m)) / (2.0 m)

F' = 980 N

F' is the force of the spring, contrary to the F which is the force on spring.

These two, however, are Third Law force pair and must have equal magnitudes.

• The compression x [based on Hooke's Law] therefore is

x = 980 N / 10000 N m^–1

x = 0. 098 m

[or 9.8 cm]

Further Physics behind

"Why isn't the weight given in this question? when obviously diving board's weight is not negligible?"

• At the beginning when the child wasn't standing at one end, the force at hinge is already adjusted together with the force on spring to deal with the weight. 
• In the above, it is initially in equilibrium with the spring having compressed already.
• Now that the child is standing on one end, the force at hinge has to increase and the force by the spring must also increase. 
• This increase in the spring's force is equal to the F [or F'] and that the corresponding additional compression is the x being asked in this question.
• In summary, you really do not need to know the weight of the diving board.