##### find |F| physics questions

label Physics
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5

A 8.7 m long elastic band is used to launch a toy glider. Once stretched to a length of 39 m, the band releases 1.5 J of elastic potential energy when you let it contract 0.063 m. What forward force is the band exerting on the toy glider?

The number of significant digits is set to 3;

May 2nd, 2015

Best Answer:The elastic band's extension is x = 39m - 8.7m = 30.3m
EPE = 0.5*k*x^2 = 1.5 J
so the band's spring constant is k = 2*1.5J / 30.1^2 = 3.31*10^-3 N/m

F = k*x
so the force on the glider is F = 3.31*10^-3 N/m * 0.063m = 2.08*10^-4 N

-----------------------

The spring's spring constant is k = 2*EPE/x^2
but F = k*x = 2*EPE / x
(here the extension x is the same in both cases = 1.5mm)

so the force is F = 2*3.0 / 1.5*10^-3 N = 4000 N

May 2nd, 2015

...
May 2nd, 2015
...
May 2nd, 2015
Oct 19th, 2017
check_circle
Mark as Final Answer
check_circle
Unmark as Final Answer
check_circle