A block with a mass of 0.5kg is pushed against a horizontal spring of negligible mass and spring constant 450 N/m until the spring is compressed a distance x. When it is released, the block travels along a frictionless, horizontal surface to the bottom of a verticle circular track of radius 1 m, and continues to move up the track.

Determine the minimum distance the spring needs to be compressed (i.e minimum x) such that the block mass will not fall off at the top of the track.

PE (of the spring) =KE (of the block)

k•x^2/2 = m•v^2/2
v^2 = k•x/m
KE(of the block) = PE (of the block)
m•v^2/2 = m•g•h
v^2 = 2•g•h =2•g•R
k•x/m = 2•g•R
x=2•g•R•m/k =2•9.8•1•0.5/450 = 0.0218 m = 2.18 cm

hi there thanks for trying to help, although my tutor has given the answer as 0.233m.

So l am therefor still puzzled by this question ... please help!!!

To solve this problem, we need to consider the forces acting on the block at the top of the track. At this point, the downward force due to gravity is balanced by the upward normal force provided by the track. The normal force will be maximum at this point.

1. Find the maximum force of static friction acting on the block at the top of the track (F_friction_max):
- The normal force (N) is equal to the weight of the block, which is given by N = m * g, where m is the mass of the block (0.5 kg) and g is the acceleration due to gravity (9.8 m/s^2).
- The maximum force of static friction (F_friction_max) is equal to the product of the coefficient of static friction (μ_s) and the normal force (N). Since the surface is frictionless, the maximum force of static friction is zero, F_friction_max = 0.

2. Determine the minimum speed the block needs at the top of the track to maintain contact with the track (v_min):
- At the top of the track, the block is moving in a circular motion. The centripetal force required to keep the block moving in a circle is supplied by the normal force (N).
- The centripetal force (F_c) is given by F_c = m * v^2 / r, where m is the mass of the block (0.5 kg), v is the speed of the block, and r is the radius of the circular track (1 m). At the point where the block is moving at the minimum speed, the centripetal force is equal to the maximum static friction force, F_c = F_friction_max = 0.
- Set up the equation F_c = 0 and solve for v_min:
m * v^2 / r = 0
v^2 = 0
v_min = 0 m/s

3. Determine the minimum distance the spring needs to be compressed (minimum x) such that the block will not fall off at the top of the track:
- The spring potential energy (PE_spring) is given by PE_spring = 0.5 * k * x^2, where k is the spring constant (450 N/m) and x is the distance the spring is compressed.
- At the minimum distance x, all of the spring potential energy will be converted into the kinetic energy of the block, given by KE_block = 0.5 * m * v_min^2 = 0.
- Set up the equation PE_spring = KE_block = 0 and solve for minimum x:
0.5 * k * x^2 = 0
x^2 = 0
x = 0 m

Therefore, the minimum distance the spring needs to be compressed (minimum x) such that the block will not fall off at the top of the track is 0 meters.

To find the minimum distance the spring needs to be compressed, we can consider the conservation of mechanical energy.

The total mechanical energy of the block-spring system is conserved during the motion. At the top of the track, the block has potential energy due to gravity and kinetic energy.

Let's break down the problem into different stages:

Stage 1: Compression of the spring
At this stage, the block is pushed against the spring until it is compressed a distance x. The work done to compress the spring can be calculated using the equation:

Work = (1/2)kx^2

where k is the spring constant and x is the distance the spring is compressed.

The work done on the block is stored as potential energy in the compressed spring.

Stage 2: Release of the block
When the block is released, it starts to move along the horizontal surface towards the bottom of the vertical circular track. Since there is no friction, the mechanical energy is conserved.

Stage 3: Motion along the vertical circular track
The block continues to move along the track due to the centripetal force provided by the normal force from the track's surface. At the top of the track, the normal force from the track is zero, and the block will start to lose contact with the track if the centripetal force is not enough to keep it moving in a circular path.

To determine the minimum distance the spring needs to be compressed, we need to calculate the velocity of the block at the top of the track and ensure that the centripetal force is greater than or equal to the gravitational force on the block.

Stage 4: Verification
We need to calculate the velocity of the block at the top of the track using the conservation of mechanical energy.

The potential energy due to gravity at the top of the track is converted into kinetic energy:

mgh = (1/2)mv^2

where m is the mass of the block, g is the acceleration due to gravity, h is the height of the track (which equals 2 times the radius of the track), and v is the velocity of the block at the top of the track.

Stage 5: Centripetal force evaluation
Now, we need to ensure that the centripetal force is greater than or equal to the gravitational force at the top of the track:

Fc = mv^2 / r

where Fc is the centripetal force, m is the mass of the block, v is the velocity at the top of the track, and r is the radius of the track.

By comparing the centripetal force to the gravitational force, we can determine if the block will remain on the track.

To find the minimum distance the spring needs to be compressed, we repeat the above steps, varying the value of x until we find the smallest value that satisfies the condition of the block not falling off the top of the track.