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;
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
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