A glider with mass m = 0.200 kg sits on a frictionless horizontal air track, con

Question

A glider with mass m = 0.200 kg sits on a frictionless horizontal air track, connected to a spring of negligible mass with force constant k = 5.00 N/m. Suppose the glider is initially at rest at x = 0, with the spring unscratched. Then you apply a constant force F^rightarrow with magnitude 0.696 N to the glider. What is the glider's speed when it has moved to x = 0.100 m? SOLUTION SET UP shows our sketch. Mechanical energy is not conserved because of the work W_other done by the force F^rightarrow, but we can still use the energy relationship of W_other = (K_f + U_grav, f + U_cl, f) = (K_f + U_grav, f + U_cl, f). Let the initial point be x = 0 and the final point be x = 0.100 m. The energy quantities are K_f = 0, k_f = 1/2 (0.200 kg)v_f^2 U_f = 0. U_f = 1/2 (5.00 N/m)(0.100 m)^2 = 0.0250 J W_other = (0.696N)(0.100 m) = 0.0696J SOLVE Putting the places into W_other = (K_f + U_grav, f + U_cl, f) - (K_f + U_grav, f + U_cl, f), we obtain K_f + U_f = K_f + U_f + W_other 1/2 (0.200 kg)v_f^2 + 0.0250 J = 0 + 0 + 0.0696 J v_f = 0.668m/s REFLECT The total mechanical energy of the system changes by an amount equal to the work W_other done by the applied force. Even though mechanical energy isn't conserved, energy considerations simplify this problem greatly. Part A - Practice Problem: If the 0 696 N force is removed when the glider roaches the 0 100 in pout, at what distance from the starting point does the glider come to rest7 Express the distance In maters to three significant figures. _______

Explanation / Answer

at x= 0.10 m

Spring elasric PE = k x^2 / 2 = 0.025 J

KE = 0.0446 J

now force is removed suppose then suppose at some position x

Spring PE = k x^2 / 2

and KE = 0

Applying energy conservation,

0.025 + 0.0446 = 5 x^2 /2 + 0

x = 0.167 m

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