## The position of a particle moving along the x axis varies in time according to the expression x = 4t 2, where x is in meters and t is in seconds. Evaluate its position at the following times. (a)t=2.10s ______m (b)t=2.10s +Δt xf=________m (c)Evaluate the limit of Δx/Δt as Δt approaches zero to find the velocity at t = 2.10 s. _______m/s …?

October 13, 2019// Blog//

**Answer:**

The spring constant of the spring is 11.65 N/m

**Explanation:**

It is given that, a 50 g mass hanger hangs motionless from a partially stretched spring. When a 65 gram mass is added to the hanger, the spring stretch increases by 10 cm.

Initial mass, m = 50 g = 0.005 kg

Final mass, m’ = 50 g + 65 g = 115 g = 0.115 kg

Initially, the elongation is x (say). The spring stretch increases by 10 cm or 0.1 m.

We know that,

**F = k x**

**Where**

**k is the spring constant**

**For the initial condition :**

………….(1)

**For the final condition :**

……..(2)

**Solving equation (1) and (2)**

**x = 0.0042 m**

Now put the value of x in equation (1)

**Hence, this is the required solution.**