A23.0 kg block at rest on a horizontal frictionless air track is connected to the wall via a spring. the equilibrium position of the mass is defined to be at x=0. somebody pushes the mass to the position x= 0.350 m, then lets go. the mass undergoes simple harmonic motion with a period of 4.10 s. what is the position of the mass 3.403 s after the mass is released?

Explanation:

We use the harmonic motion position equation:

where A = 0.350 and for t = 0

so:

and also:

so we have:

x(t)=0.350cos(1.532 t)

For t = 3.403 s

x(3.403)=0.350cos(1.532 (3.403)) = 0.348 m

kepler used the data from tycho blahe (his mentor) to construct the laws of planetary motion.

dec 23, 2016 - a load of bricks weighing 680 n a distance of 10 m across a parking lot. if if the worker pushes the cart with a constant force of 209 n, what work= force x distance = 220 n x 10 m = 2200 n.m = 2200 j i don't know where the person got it from, but the initial part, work= force x distance me.