A 1.150 kg air-track glider is attached to each end of the track by two coil springs. It takes a horizontal force of 0.900 N to displace the glider to a new equilibrium position, x= 0.170 m. Find the effective spring constant of the system. Submit Answer Tries 0/12 The glider is now released from rest at x= 0.170 m. Find the maximum x-acceleration of the glider. Submit Answer Tries 0/12 Find the x-coordinate of the glider at time t= 0.470T, where T is the period of the oscillation. Submit Answer Tries 0/12 Find the kinetic energy of the glider at x=0.00 m.

k = 5.29

a = 0.78m/s²

KE = 0.0765J

Explanation:

Given-

Mass of air tracker, m = 1.15kg

Force, F = 0.9N

distance, x = 0.17m

(a) Effective spring constant, k = ?

Force = kx

0.9 = k X0.17

k = 5.29

(b) Maximum acceleration, m = ?

We know,

Force = ma

0.9N = 1.15 X a

a = 0.78 m/s²

c) kinetic energy, KE of the glider at x = 0.00 m.

The work done as the glider was moved = Average force * distance

This work is converted into kinetic energy when the block is released. The maximum kinetic energy occurs when the glider has moved 0.17m back to position x = 0  

As the glider is moved 0.17m, the average force = ½ * (0 + 0.9)

Work = Kinetic energy

KE = 0.450 * 0.17

KE = 0.0765J

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explanation:

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards