Question

A horizontal block-spring system with the block on a frictionless surface has total mechanical energy E = 53.9 J and a maximum displacement from equilibrium of 0.197 m.

(a) What is the spring constant?
N/m

(b) What is the kinetic energy of the system at the equilibrium point?
J

(c) If the maximum speed of the block is 3.45 m/s, what is its mass?
kg

(d) What is the speed of the block when its displacement is 0.160 m?
m/s

(e) Find the kinetic energy of the block at x = 0.160 m.
J

(f) Find the potential energy stored in the spring when x = 0.160 m.
J

(g) Suppose the same system is released from rest at x = 0.197 m on a rough surface so that it loses 12.2 J by the time it reaches its first turning point (after passing equilibrium at x = 0). What is its position at that instant?
m

Answer 1

Given is :-

Total mechanical energy E = 53.9 J

Maximum displacement = 0.197 m

Now,

part - a

The spring constant is given by

which gives us

Part - b

The kinetic energy of the system at the equilibrium point is equal to the total energy, thus

Part - c

Maximum speed of the block is = 3. 45 m/s

The mass is given by

or

which gives us

Part - d

The speed of the block at 0.160m displacement is

by plugging the values we get

which gives us

Part -e

The kinetic energy at x = 0.160 m is

which gives us