Spring Force and Work Question?

Question

A 200-g block is pressed against a spring of force constant 1.40 kN/m until the
block compresses the spring 10.0 cm. The spring rests at the bottom of a ramp inclined at
60.0° to the horizontal. Using energy considerations, determine how far up the incline the
block moves before it stops (a) if there is no friction between the block and the ramp and
(b) if the coefficient of kinetic friction is 0.400.

Answer 1

The energy stored in the compressed spring = 1/2kx^2 where k is the spring constant and x is the compression.
E = 1/2 x 1400 x 0.1 x 0.1
E = 7 Joules
(a) without friction, all of this energy will be finally converted to Potential energy of the block
7 = mgh
7 = 0.2 x 9.8 x h
h =3.571 m
now the distance, d along the incline will be
d = h/sin60
d = 3.571/sin60
d = 4.124m

(b) In this case, the energy stored in the spring will be converted to PE of block + work done to overcome friction
Work done against friction = F xd
= umgcos60 x d
= 0.4 x 0.2 x 9.8 x 0.5 x d
=0.392d..... (1)
PE = mgh
= 0.2 x 9.8 x dsin60
=1.697d.....(2)
Now the total energy = (1) + (2)
7 = 0.392d + 1.697d
7 = 2.089d
d= 3.35m

Answer 2

There are three forms of energy in place: kinetic (KE), gravitational potential (GPE), and elastic potential (EPE). KE = 0.5*mv^2, GPE = mgh, and EPE = 0.5*kx^2. Due to conservation of energy, they must have a constant sum (in part a) at least, but I'll get to part b) later).

Use the compressed position of the spring as height h = 0. GPE = mgh = mg(0) = 0. If the block moves a distance d up the ramp, it achieves a height of h = d*sin(60) and GPE = mgh = mgd*sin(60).

When compressed against the spring, the block is not moving. Thus, initially, v = 0 and KE = 0.5*mv^2 = 0.5*m(0) = 0. When the block reaches its maximum height, it stops there as well before falling back down, so once again KE = 0.

We know that the intial compression of the spring is 10 cm = 0.1 m, so initially EPE = 0.5*kx^2 = 0.5(1400 N/m)(0.1 m)^2. After the block has launched, the spring is no longer compressed, so x = 0 and EPE = 0.5*kx^2 = 0.5k(0) = 0.

Solving (KE + GPE + EPE)_initial = (KE + GPE + EPE)_final should not be difficult now (considering that KE_initial, KE_final, GPE_initial, and EPE_final are all equal to zero), and d will be your only unknown.

To solve with friction, you must instead solve (KE + GPE + EPE)_initial = (KE + GPE + EPE)_final - WOF, where WOF is the work of friction and is equal to umg*cos(60)*d. u is the coefficient of kinetic friction, and mg*cos(60) is the normal force of the ramp against the block on the incline, so umg*cos(60) is the force of friction. d is the path length along with the force of friction acts, so their product is the work done by friction. The equation has another term in it this time, but d is still the only unknown.